Reinforced Concrete Design-PILLAI & MENON-FULL

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. .' .. About the Authors '

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Dr. S. Unnikrishria Pillai ("SITHARA", 3 1 - Srinagar Lane, Thiruvananthapuram .his Engineering education ~t the University of Kerala and 695025, Keralaj-had , .; Queen's Univhwty, Canada. He has taugHt - at College o f Engineering, Thiruvnnanthapuram; Queen's University, Canada; University o f Sulaimaniya, Iraq; ~ o ~ a l : M i l ' i t a r ~ ~ C o l of le@ Canada and REC, Calicut. His major research interests have.bee11in the areasof Limit States Design o f concrete structures, Strength and Stability of steel Beam-Columns and Rehabilitation o f structures. He also has wide experience in design and construction o f various RC Structures. Dr. Pillai has published over 45 technical papers, authored the book Reinforced Concrete Design (McGraw-Hill Ryerson, Canada), co-authored Ch 8 - BeamColmnns of SSRC G~iideto Stabiliy Design Criteria for Metal Structwes (Jolm Wiley & Sons. New York) and contributed to the design provisions for Beam~ol;;mns in thc Canadian Standard S16.1: Steel Structures for Buildings (Limit Smes Desfgn). He has received Several awards, including Kerala University Scholarship for First Rank, Canadian Commonwealth Scholarship, Architectural Engineering Division Gold Medal (I.E., India), Sir Arthur Conon Memorial Prize (I.E., India), UP Government National Award, and Fellowship of NRC, Canada. '

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D r Devdas Menon had his civil engineering education at I I T Madras and I I T Delhi. Soon after graduation, he worked as a structural engineer at New Delhi, and later took to academics. He is presently a Professor in the Department o f Civil. ~ngineeringat I I T Madras. Prof. Menon has authored numerous technical publications and has received awards related to research and teaching. His current research interests in Engineering are mainly in Reinforced and Prestressed Concrete Design (especially bridges, chimneysand water tanks) and Structural Reliability. The research is largely directed towards solving practical problems and improving national codes and standards. Prof. Menon promotes a holistic approach to education, in which "awakening" has a central role to play. He is the author of a book titled Stop sleep-walking through lift.! published by Yogi Impressions in 2004. For more information, visit www.devdasmenon.coni.

S Unnikrishna Billai Fellow. American Sociefv of Civil Enuineers Director, Cooperative Academy of Professional ~ d i c a t i o n~, r i v k d r u m (Formerly Principal, Regional Engineering College, Calicut) [Email: supillai&snl.com]

Devdas Menon Professor, Department of Civil Engineering Indian Institute of Technology, Madras, Chennai 600 036. [Email: [email protected]]

Cover Feature: Engineering Design 8 Research Centre, LARSEN 8 TOUBRO LTD, ECC Division. Chennai - an fib award-winning concrete structure (8 686 m2 area), shaped like a tree with two wings, with the first floor suppo~tedon rib beams emanating from a central column, and each subsequent floor rotated in plan by 60 degrees. Architects: K.S. Ranganath, Chennai. Structural Consultants: EDRC, L&T Ltd, ECC Division. Chennai

Twta McGraw-Hill Publishing Con~panyLimited

Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw-Hill and its authors are supplying information but are- not attempting to render engineering or other professional' services. If such services are required, the assistance of an appropriate professional should be sought.

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i Tata McGraw-Hill

O 2003, 1998, Tata McGraw-Hill Publishing Company Limited Fifth reprint 2005 KLLYCRBBRABDl

N o part of this publication can be reproduced in any form or by any means without the prior written permission of the publishers This edition can be exported from India only by the publishers, Tata McGraw-Hill Publishing Company Limited ISBN 0-07-049504-1 Published by Tata McGraw-Hill Publishing Company Limited, 7 West Patel Nagar, New Delhi 110008, typeset in Times New Roman a t Indian Institute of Technology, Madras, Chennai 600036 and Gopaljee Enterprises, Delhi l I0 053 Cover: Mudrak

This second edition of Reinforced Concrete Design has been brought out incorporating the recent revisions in IS 456 : 2000 and related codes, latest developments in the field and certain additional topics suggested by users of the first edition. Thc contents of the first edition have been thoroughly reviewed and updated, the examples and answers to problems reworked to incorporate the many changes in the revised codes, and review questions and references expanded and updated. Apart from the revision of the sixteen chapters of the first edition, a new chapter (Chp 17: Special Selected Topics) has been added. Major topics added include compression field theory, seut-cudtie model, interface shear and shear-jriction for shear design; shear connectors; fire resistance design; cracking under direct tension and thermal and shrinkage cracking; and design of cocmterfort retaining walls. With these additions, this book incorporates all the topics in reinforced concrete design generally required for a civil engineering degree programme in Indian universities. The second edition has been entirely reformatted, with several figures revised and new figures added for improved clarity. However, the unique fcatures of the first edition relating to coverage and planning of contents, description and illustration, presentation and overall organization of the text have been retained. So also the exceptional chapters on structural systems, good detailing and consrructionpractices, and earthquake resistant design, as well as the design aids, which made the book stand apart, have all been expanded and retained. With these and the additional special selected topics mentioned carlier (all of practical relevance), this book should also be of use to postgraduate students, teachers and practicing engineers. This book lays great emphasis on conceptual clarity and strcngth in fundamentals. The student is encouraged to raise questions, to relate to field experience, to develop a 'structural scnse', to appreciate proper 'documentation' and 'detailing', to analyse results and to synthesize knowledge. A set of stimulating 'review questions' is posed at the end of each chapter, in addition lo a large number of illustrative workcd examples. An exhaustive set of problems (with answers for 'analysis' type problems)

is also included at the end of each chapter. Extensive chapter-wise I.eferences are also listed, to enable the research-minded reader to pursue further study. , The authorswish to acknowledge gratefully the assistance rendered by Mr Battu Sri Rama Krishna, Mr Sanjay K Nayak, Mr Aby Abraham and Mr Prathish K Unni of IIT Madras in incorporating some of the changes in this edition. The authors also express their gratitude to colleagues,. students, friends and family members, who contributed in making this bookpossible. The authors welcome suggestions from readers for improving this book in any manner.

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This book is inspired by an earlier book, also entitled Reinforced Concrete Design, authored by S UPillai and D W Kirk, and publisl~ed by McGraw-Hill Ryerson (Canada) in 1983, with a second edition in 1986. The success of that book in Canada had pmnptcd McGraw-Hill (Singapore) to suggest the launching of an international edition suitable for use in Asian countries. This was found diificult, because reinforced concrete design (unlike other subjects such as structural analysis) is highly dependent on building codes, which differ from country to country. Some academics in India, familiar with the earlier book (based on Canadian Standards), suggested that a similar book based exclusively on Indian Standards would be welcome in India. Tata McGraw-Hill (New Delhi) also readily agreed. to do the publishing. This book is an ourcome of these promptings. Altho~lghthe basic format of the earlier book was used as a guideline, thc contents have been tl~oroughlyreorganised, expanded and written afresh. This book has sixteen chapters, which cover, or rather, uncover (!) all the fundamental topics in reinforced concrete design, generally taught in a first course in the B.Tech. (Civil Engineering) curriculum in Indian universities. Two spccial topics, of practical relevance, included in this book are Good Dcruilirzg and Construction Practices and Specin1 P,avi,~ionsfor. E~~rrlzq~roke-Rcsi~'fn,~f Design. The various topics have been discussed i n depth, and the coda1 pmvisions of IS 456 : 1978 havc bee11 analysetl critically and compared with foreign codes, whercver iclevant. This book, therefore, sllould also bc of usc to postgraduate stutl~nts, teachcrs and practising engineers. The modcrn philosophy o i h i t stores design is followed in this book - in kceping with the current design practice, both in India and abroad. The use of the traditional workh~gsrfrss ,,rrrlzod is lirnitcd to nnalysis of flcxural mcmbers under service loads, which is required for investigating the limit states of serviceability (deflection, cracki~~g). This hook is not a conventiol~alexamination-oriented textbook, although it does contain the neccssary infomation (including a large of ~iumber of illustrative

Ylll PREFACE TO THE

FIRST

EDITION

examples) required to face examinations. The emphasis here is on conceptuul clarity and strength in fundamentals. The student is encouraged to raise questions, to relate to field experience, to develop a 'structural sense', to appreciate proper 'detdling', to aqaiyse results, and to synthesise knowledge. A set of stimulating 'review questions' is posed at the end of each chapter. A fairly exhaustive set of problems(wit11 answers for 'analysis' type problems) is also included at the end of each chapter. Extensive chapterwise references are also listed, to enable the rcsea~cli-mindedrender to pi~rsue further study. After gaining a proper understanding of design applications based on first principles, the student is encouraged to make use of time-saving 'analysis aids' and 'design aids', in the fonn of appropriate tables, charts and diagrams. Indeed, this is what practising designers invariably resort to. Some typical 'design aids' have been derived and included in this book, wherever relevant. The relevant algorithms have also bee11 explained, to facilitate the making of similar design aids by the softwareoriented student. Readers we also encouraged to make ose of SP 1 6 : 1980 (published by the Bureau of Indian Standards) for this purpose. This book has borrowcd extensively from the book authored by S U Pillai and D WKirk, rcfel~edto earlier, and this fact is gratefully acknowledged. The authors also acknowledge their debt to various agencies, particularly the Bureau of Indian Standards, for their published material to which refemnces are made in this book. A special word of thanks is due to Beljith P., for having done an excellent job in preparing the typescript for this book. The authors also exprcss their gratit~~de to all, including colleagues, shldents, friends and family members, who contributed in making this book possible. The authors welcome suggestions from readers for improving this book in any manner. This book is dedicated to students of civillstructural engineering, with the hope that they will find the learning of reinforced concrete design a rewarding experience.

S U N N ~ S H NPILLAI A DEVDAS MENON

Preface to t h e Second Edition

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Preface to t h e First Edition 1. REINFORCED CONCRETE STRUCTURES 1.1 lntroduction 1 1.2 Plain a n d Reinforced C o n c r e t e 4 1.2.1 Plain Concrete 4 1.2.2 Reinforced Concrete 5 1.3 Objectives o f Structural Design 7 1.4 Reinforced C o n c r e t e Construction 8 1.5 Structural S y s t e m s 9 1.6 Reinforced C o n c r e t e Buildings 9 1.6.1 Floor Systems 11 1,6.2 Vertical Framing System 1 7 1.6.3 Lateral Load Resisting Systems 19 1.7 Structural Analysis a n d Design 21

1.8 Design C o d e s a n d H a n d b o o k s 22 1&I Purpose of Codes 22 1.8.2 Basic Code for Design 22 1.8.3 Loading Standards 23 1.8.4 Design Handbooks 23 1.8.5 Other Related Codes 23 Review Q u e s t i o n s 24 References 24

2. BASIC MATERIAL PROPERTIES 2.1 Introduction 25 2.1.1 Concrete Technology 25

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CONTENTS

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2.2 Cement 26 2.2.1 portland Cements 26 2.2.2 Other Cements 28 .:.2:2.3 Tests on Cements 29 2.3 ~ g g r e g a t e29 2.3.1 Aggregate Properties and Tests 30 2.3.2 Grading Requirements of Aggregate 31 2.4 Water 32 2.4.1 Water Content and Workability of Concrete 33 2.4.2 WaterCernent Ratio and Strength 34 2.4.3 Water for Curing 35 2.5 Admixtures 36 2.5.1 Types of Chemical Admixtures 37 2.5.2 Types of Mineral Admixtures 37 2.6 Grade o f Concrete 38 2.6.1 Characteristic Strength 39 2.7 Concrete M i x Design 40 2.7.1 Nominal Mix Concrete 40 2.7.2 Design Mix Concrete 41 2.8 Behaviour o f Concrete u n d e r Uniaxial Compression 42 2.8.1 influence of Size of Test Specimen 43 2.82 Stress-Strain Curves 45 2.8.3 Modulus of Elasticitv and Poisson's Ratio 46 2 8 4 *nl#Jenceof D,ml on ot ~ o a a m gon Slross-Slra n C u ~ e49 m ve Stress of Concrete t i Des qn Pracwe 50 2 8 5 M a x ~ m ~Compress 2.9 Behaviour o f Concrete u n d e r Tension 50 2.9.1 Modulus of Rupture 51 2.9.2 Splitting Tensile Strength 52 2.9.3 Stress-Strain Curve of Concrete in Tension 52 2.9.4 Shear Strength and Tensile Strength 53 2.10 Behaviour of Concrete u n d e r C o m b i n e d Stresses 53 2.10.1 Biaxial State of Stress 53 2 . 0 2 Influence of Shear Stress 53 2.10.3 Behaviour under Triaxiai Compression 55 2.11 Creep o f Concrete 55 2.1 1.I Time-Dependent Behaviour under Sustained Loading 55 2.11.2 Effects of Creep 56 2.11.3 Factors Influencing Creep 57 2.11.4 Creep Coefficient for Design 57

.2.12 Shrinkage a n d Temperature Effects in Concrete 57 2.121 Shrinkage 57 2.12.2 Temperature Effects 59 2.13 Durability o f Concrete 59 2.13.1 Environmental Exposure Conditions and Code Requirements 61 2.13.2 Permeability of Concrete 63

2.13.3 Chemical Attack on Concrete 63 2.i3.4 Corrosion of Reinforcing Steel 65 2.14 Reinforclng Steel 65 2.1.4.1 Types, Sizes and Grades 66 2.14.2 Stress-Strain Curves 67 2.15 L i s t o f Relevant i n d i a n Standards 70 Review Questions 72 References 74

3. BASIC DESIGN CONCEPTS 3.1 l n t r o d u c t l o n 77 3.1 .I Design Considerations 77 3.i.2 Design Philosophies 78 3.2 W o r k i n g Stress M e t h o d (WSM) 79

3.3 Ultimate L o a d M e t h o d (ULM) 80 3.4 Probabilistic Analysis a n d Design 80 3.4.1 Uncertainties in Design 80 3.4.2 Classical Reliability Models 82 3.4.3 Reliability Analysis and Design 84, 3.4.4 Levels of Reliability Methods 84 3.5 L i m i t States M e t h o d (LSM) 85 3.5.1 Limit States 85 3.5.2 Multiple Safety Factor Formats 85 3.5.3 Load and Resistance Factor Design Format 86 3.5.4 Partial Safety Factor Format 86 3.6 C o d e Recommendations f o r L i m i t States Design 87 3.6.1 Characteristic Strengths and Loads 87 3.6.2 Partial Safety Factors for Materials 88 3.6.3 Parliai Safety Factors for Loads 88 3.6.4 .Design Stress-Strain Curve for ConCrete 89 3.6.5 Design Stress-Strain Curve for Reinforcing Steel 90 Revlew Questions 93 References 94

4. BEHAVIOUR IN FLEXURE 4.1 I n t r o d u c t i o n 95 4.1.1 Two Kinds of Problems: Analysis and Design 95 4.1.2 Bending Moments in Beams from Structural Analysis 96 4.1.3 From Bending Moment to Flexural Stresses 97 4.2 Theory o f Flexure f o r H o m o g e n e o u s Materials 97 4.2.1 Fundamental Assumption 97 4.2.2 Distribution of Stresses 97

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4.2.3 Linear Elastic Material 99 4.3 Linear Elastic k n a l y s i s o f Composite Sections 4.3.1 Distribution of Strains and Stresses 100 4.3.2 Concept of 'Transformed Section' 101 4.4 Modular Ratio a n d Cracking Moment 101 4.4.1 Modular Ratio in Reinforced Concrete 101 4.4.2 Transformed Area of ReinforcingSteel 102 4.4.3 Cracking Moment 103

CONTENTS

99

4.5 Flexural Behaviour of Reinforced Concrete 105 4.5.1 Uncracked Phase 107 4.5.2 Linear Elastic Cracked Phase 107 4.5.3 Stages Leading to Limit State of Collapse 108

4.6 Analysis a t Service L o a d s (WSM) 112 4.6.1 4.6.2 4.6 3 4 6.4 465

Stresses in Singly Reinforced Rectangular Sections 112 Permissible Stresses 115 A,ovrah.e Ben0 nQMoment 116 And ys s of Sinyl/ I7c nforcod Flanged Se21 uns 122 A#~alysts 01 DUI.nIy Heinlorced SCC~IOIIS 128 4.7 Analysis a t Ultimate Limit State 134 4.7.1 Assumptions in Analysis 134 4.7.2 Limiting Depth of Neutral +is 135 4.7.3 Analysis of Singly RetnforcbdRectangular Sections !37 4.7.4 Analysis of Singly Reinforced Flanged Sections 147 4.7.5 Analysis of Doubly Reinforced Sections 153 4.7.6 Balanced Doubly Reinforced Sections 159

4.8 Analysis of Siabs a s Rectangular Beams 160 4.8.1 Transverse Moments in One Way Siabs 161 Review Questions

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164 References 167 Problems

5. DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 169

5.1 Introduction 159 5.2 Requirements of Flexural Reinforcement 170 5.2.1 Concrete Cover 170 5.2.2 Spacing of Reinforcing Bars 172 5.2.3 Minimum and Maximum Areas of Fiexural Reinforcement 174 5.3 Requirements for Deflection Control 176 5 3.1 1)eI.ccl.on Cumro. uy ~ ' n l ' l ~ n Span g Depth Rnlos 17C 5 3 2 Code Necom!nt?noat~ons lor SpmiEfI~.ct;vcDnpll, Rxllos 177 5.4 Guidelines for Selection of Member Sizes 179 5.4.1 General Guidelines for Beam Sizes 179 5.4.2 General Guidelines for Slab Thicknesses 180

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5.4.3 Deep Beams and Slender Beams 160

5.5 Design o f Singly Reinforced Rectangular S e c t i o n s 181 5 5 1 F Xuy D mens,onS of Rectang~larSc~llons182 5 5 2 OctwnlnlnQ k e a ol Tons on Stccl 183 5 5 3 Dcs gn Cnecd lor Strcngtt! aou Deflecllon Contro 185

5.6 Design o f Continuous One-way Slabs 189 5.6.1 Simplified Structural Analysis - Use of Moment Coefficients' 190 5.6.2 Design Procedure 192

5.7 Design of ~ o u b l Reinforced y Rectangular S e c t i o n s 197 5.7.1 Design Formulas 197 5.7.2 Design Procedure for Given Mu 199

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5.8 Design of Flanged Beam Sections 203 5.8.1 Transverse Reinforcement in Flange 203 5.8.2 Design Procedure 204 5.9 Curtailment o f Fiexural Tension Reinforcement 5.9.1 Theoretical Bar Cut-off Points 210 5.9.2 Restrictions on Theoretical Bar Cut-off Points 212 5.9.3 Code Requirements 214 5.9.4 Bending of Bars 219 Review Questlons 221

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222 References 223 Problems

6. DESIGN FOR SHEAR 6.1 lntroductlon 225 6.2 Shear Stresses in Homogeneous Rectangular B e a m s 226 6.3 Behaviour of Reinforced Concrete under Shear 228 6.3.1 Modes of Cracking 228 6.3.2 Shear Transfer Mechanisms 230 6.3.3 Shear Failure Modes 232

6.4 Nominal Shear Stress 234 6.4.1 Members with Uniform Depth 234 6.4.2 Members with Varying Depth 234 6.5 Critical Sections f o r Shear Design

236 6.6 Design Shear Strength without Shear Reinforcement 238 6.6.1 Design Shear Strength of Concrete in Beams 238 6.6.2 Design Shear Strength of Concrete in Slabs 240 6.6.3 Influence of Axial Force on Design Shear Strength 241

6.7 Design Shear Strength w i t h Shear Reinforcement 242 6.7.1 Types of Shear Reinforcement242 6.7.2 Factors Contributing to Ultimate Shear Resistance 243 6.7.3 Limiting Ultimate Shear Resistance 244

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XIV CONTENTS

8 2 1 Effect of Flexural Cracking on Flexural Bond Stress 298

6.7.4 Shear Resistance of Web Reinforcement 245 6 7.5 Influence of shear on longitudinal reinforcement 247 6 7.6 Mlnimum Stirrup Relnforcement 249

8.3

Anchorage (Development) Bond 299 8 3.1 Development Length 300

8.4 Bond Failure and Bond Strength 301

6.8 Additional Comments on Shear Reinforcement Design 249

8.4.1 Bond Failure Mechanisms 301 8.4.2 Bond Tests 303 8.4.3 Factors Influencing Bond Strength 305

6.9 lnterface shear and Shear Friction 251 6.9.1 Shear frlction 251

8.5 Review of Code Requirements for Bond 305

6.9.2 Recommendation for Interface Shear Transfer 254

6.5.1 Flexural Bond 305 8.5.2 Development (Anchorage) Bond 306 8.5.3 Bends, Hooks and Mechanical Anchorages 306

6.10 Shear Connectors in Flexural Members 256 6 10.1 Shear along Horlzontai Planes 256

6.11 Shear Design Examples

- Conventional Method 257

8.6 Splicing of Relnforcement 308 8.6.1 Lap Splices 308 8.6.2 Welded Splices and Mechanical Connections 310

Review Questions 263 Problems 264

8.7 Design Examples 311

References 266

Review Questions 314

7. DESIGN FOR TORSION 7.1 Introduction 267 7.2 Equilibrium Torsion a n d Compatibility Torsion 267 7.2.1 Equilibrium Torsion 268 7.2.2 Compatibility Torsion 268 7.2.3 Estimation of Torsional Stiffness 270 7.3 General Behaviour in Torsion 271 7.3.1 Behaviour of Plain Concrete 271 7.3.2 Behaviour of Concrete with Torsional Reinforcement 273 7.4 Design Strength in Torsion 274 7.4.1 Dos gn Torsional Strengln witnodl Torsion Reinforcement 274 7.4.2 Des gn Torsional Strengln witn Tors on Reinforcement 277 7.4.3 Design Strengln (nTorsion Comb'ned wiln Flexure 280 7.4.4 Design Strenqth in Torsion Combined with Shear 282

7.5 Analysis and Design Examples 284 Review Questions 291 Problems 292 References 294

8. DESIGN FOR BOND 8.1 lntroduction 295 8.1.1 Mechanisms of Bond Resistance 295 8.1.2 Bond Stress 296 8.1.3 Two Types of Bond 296

8.2 Flexural B o n d 297

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Problems 315 References 316

9. ANALYSIS FOR DESIGN MOMENTS IN CONTINUOUS SYSTEMS

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9.1 lntroduction 317 9.1.1 Approximations in Structural Analysis 317 9.1.2 Factored Moments from Elastic Analysis and Moment Redistribution 320 9.2 Gravltv Load Patterns for Maximum Design - Moments 321 9.2.1 ~ ej sn tiomcnls n Besms 322 9.2.2 DCSgn Momelllr, .n Colbmns 323 ~

9.3 Simplified (Approximate) Methods of Analysis 324 9.3.1 Moment Coefficients for Continuous Beams under Gravity Loads 324 9.3.2 Substitute Frame Method of Frame Analysis for Gravity Loads 324 9.3.3 Simplified Methods for Lateral Load Analysis 327

9.4 Proportioning of Member Sizes for Preliminary Design 328 9.5 Estimation of Stiffnesses of Frame Elements 330 9.6 Adjustment of Design Moments at Beam-Column Junctions 331 9.7 Inelastic Analysis and Moment Redistribution 334 9.7.1 Limit Analysis 334 9.7.2 Moment Redistribution 337 9.7.3 Code Recommendations for Moment Redistr~bution341

9.8 Design Examples 345 Review Questions 353

XVI CONTENTS

CONTENTS

Problems 353 References 355

10.

SERVICEABILITY

, LIMIT STATES: DEFLECTION AND

CRACKING

357

10.1 Introduction 357 10.2 Serviceability L i m i t States: Deflection 358 10.2.1 Deflection Limits 358 10.2.2 Difficulties in Accurate Prediction of Deflections 359 10.3 Short-Term Deflections 360 10.3.1 Deflections by Elastic Theory 360 10.3.2 Effective Flexural Rigidity 361 10.3.3 Tension Stiffening Effect 362 10.3.4 Effective Second Moment of Area 364 10.3.5 Average Ian for Continuous Spans 366 10.3.6 Effective Curvature Formulation 368 10.3.7 Additional Sholl-Term Deflection due to Live Loads alone 373 10.4 Long-Term Defiection 380 10.4.1 Defiection Due to Differential Shrinkage 381 10.4.2 Deflection Due to Creep 384 10.4.3 Deflection Due to Temperature Effects 387 10.4.4 Checks on Total Deflection 388 10.5 Serviceability L i m i t State: Cracking 391 10.5.1 Cracking in Reinforced Concrete Members 391 10.5.2 Limits on Cracking 393 10.5.3 Factors Influencing Crackwidths 393 10.5.4 Estimation of Flexural Crackwidth 395 10.5.5 Estimation of Crackwidth under Direct and Eccentric Tension 405 10.5.6 Thermal and Shrinkage Cracking 409 Review Questions 412

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u n d e r Gravity L o a d s 460 11.4.1 Codal Procedures Based on the Equivalent Frame Concept 460 11.4.2 Proporlioning of Slab Thickness, Drop Panel and Column Head 463 11.4.3 Transfer of Shear and Moments to Columns in Beamless Two-Way Siabs 467 11.5 Direct Design M e t h o d 469 11.5.1 Limitations 469 11.52 Told Dnslgn Mome.its for a Spnn 170 11 5.3 Long.l..n,nal D s ~ b r A o n ot Told Des gn Momcm 470 11.5.4 Apponioning of Moments to M do e Stlips, Co Jmn Strips atla Beams

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11.5.5 Loads on the Edge Beam 476 11.5.6 Torsion in the Edge Beam 476 11.5.7 Moments in Columns and Pattern Loading 476 11.5.8 Beam Shears in Two Way Slab Systems with Flexible Beams 480 11.6 Equivalent Frame M e t h o d 481 1 1 .6.l Equivalent Frame for Analysis 481 1 1 6.2 Slab-Beam Member 483 11.6.3 Loading Patterns 491 11.6.4 Design Moments in Slab-Beam Members 492 1 1 6 5 Design Moments in Columns and Torsion in Transverse Beam 494

11.7 Reinforcement Details i,n Column-Supported T w o - W a y S l a b s 494

References 415

11.1 Introduction 417 I . . One-way and Two-Way Actions of Slabs 417 11.1.2 Torsion in Two-Way Slabs 419 11.1.3 Difference Between Wail-Supported Siabs and BeamICoiumn Supported Slabs 420 11.2 Design o f Wall-Supported Two-Way Slabs 422 112.1 Slab Thickness Based on Deflection control Criterion 422 11.2.2 Methods of Analysis 422 11.2.3 Uniformly Loaded and Simply Supported Rectangular Slabs 423

11.2.4 Uniformly Loaded 'Restrained' Rectangular Slabs 427 11.2.5 Shear Forces in Uniformly Loaded Two-Way Siabs 435 11.2.6 Design of Circular, Triangular and other Slabs 448 11.2.7 Two-Way Slabs Subjected to Concentrated Loads 454 11.3 Design o f Beam-Supported Two- Way Slabs 454 1 1 .3.l Behaviour of Beam-Supported Siabs 454 113.2 Use of Codal Moment Coefficients for Slabs Supported on Stiff Beams 454 11.3.3 Slabs Supported on Flexible Beams - Codal Limitatipns 456 11.3.4 The 'Equivalent Frame' Concept 456

11.4 Design of Column-Supported Slabs ( w i t h i w i t h o u t Beams)

Problems 413

11. DESIGN OF T W O - W A Y S L A B SYSTEMS

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417

11.8 Shear in Column-Supported Two- Way Slabs 497 11.8.1 One-way Shear or Beam Shear 497 11.8.2 Two-Way Shear or Punching Shear 499 11.9 D e s i g n Examples o f Column-Supported T w o - w a y Slabs 504 Review Q u e s t i o n s 528 Problems 529 References 531

XVlll CONTENTS CONTENTS

12. DESIGN OF STAIRCASES

533

12.1 Introduction 533 12.2 Types of Staircases 535 12.2.1 Geometrical Configurations 535 12.2.2 Structural Ciassification 536 12.3 Loads and Load Effects o n Stair Slabs 540 12.3.1 Dead Loads 541 12.3.2 Live Loads 541 12.3.3 Distribution of Gravity Loads in Special Cases 541 12.3.4 Load Effectsin isolated Tread Siabs 542 12.3.5 Load Effectsin Waist Siabs 542 12.3.6 Load Effects in Tread-Riser Stairs 544 12.4 Deslgn Examples of Stair Slabs Spanning Transversely 547

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Review Questions 562

13.1 Introduction 565 13.1.1 Ciassification of Columns Based on TvDe of Reinforcement 565 c _onding 5GG 13 1.2 Classlf,cationof ColJm!~sBased on ~ j p 01 13.1.3 Ciassitculon of Col.nlns Based on S enacnness Ratios 5ti8 13.2 Estimation o f Effective Length of a Column 569 13.2.1 Definition of EffectiveLength 569 13.2.2 Effective Length Ratios for ideaiised Boundary Conditions 570 13.2.3 Effective Length Ratios of Columns in Frames 573 13.3 Code Requirements o n Slenderness Limits, Minimum Eccentricities a n d Reinforcement 581 13.3.1 Slenderness Limits 581 13.3.2 Minimum Eccentricities 582 13.3.3 Code Requirements on Reinforcement and Detailing 582 13.4 Design of Short Columns under Axial Compression 586 13.4.1 Conditions of Axial Loading 586 13.4.2 Behaviour under Service Loads 587, 13.4.3 Behaviour under Uitimate Loads 588 13.4.4 Design Strength of Axially Loaded Short Columns 590 13.5 Design of Short Columns under Compression with Uniaxial Bending 594 13.5.1 Distribution of Strains at Uitimate Limit State 594 13.5.2 Modes of Failure in Eccentric Compression 596 13.5.3 Design Strength: Axiai Load-Moment Interaction 597 13.5.4 Analysis for Design Strength 600

Review Questlons 649 Problems ,650

14. DESIGN OF FOOTINGS AND RETAINING WALLS C

References 564

13. DESIGN OF COMPRESSION MEMBERS

13.5.5 Use of Interaction Diagram as an Analysis Aid 610 13.5.6 Non-dimensional interaction Diagrams as Design Aids 618 13.6 Deslgn of Short Columns under Axial Compression w i t h Blaxial Bendlng 625 13.6.1 Biaxiai Eccentricities 625 13.6.2 lnteraction Surface for a Biaxiaiiy Loaded Coiumn 627 13.6.3 Code Procedure for Design of Biaxialiy Loaded Columns 629 13.7 Design of Slender Columns 634 13.7.1 Behaviour of Siender Coiurnns 634 13.7.2 Second-Order Structurai Analysis of Siender Coiumn Structures 639 13.7.3 Code Procedures for Design of Siender Coiumns 639

References 653

12.5 Deslgn Examples of Stair Slabs Spanning Longitudinally 552 Probiems 562

565

XIX

14.1

655

lntroductlon 655

14.2 Types of Footlngs 656 14.2.1 isolated Footings 658 14.2.2 Combined Footings 658 14.2.3 Wail Footings 659 14.3 Soil Pressures under Isolated Footings 659 14.3.1 Allowable Soil Pressure 659 14.3.2 Distribution of Base Pressure 660 14.3.3 instability Probiems: Ovedurning and Sliding 664 14.4 General Deslgn Considerations and Code Requirements 665 14.4.1 Factored Soil Pressure at Uitimate Limit State 665 14.4.2 Generai Deslgn Considerations 667 14.4.3 Thickness of Footing Base Slab 667 14.4.4 Design for Shear 667 14.4.5 Design for Flexure 669 14.4.6 Transfer of Forces at Coiumn Base 671 14.4.7 Plain Concrete Footings 673 14.5 Design Examples o f Isolated and Wall Footings 674 14.6 Design of Comblned Footlngs 692 14.6.1 Generai 692 14.6.2 Distribution of Soil Pressure 693 14.6.3 Geometry of TwoGolumn Combined Footings 693 14.6.4 Design Considerations in Two-Column Footings 693 14.7

Types of Retalnlng Wails and Thelr Behaviour 703

14.8 Earth Pressures a n d Stability Requirements 706 14.8.1 Lateral Earth Pressures 706 14.8.2 Effect of Surcharge on a Level Backfill 708

CONTENTS

XX CONTENTS

14.8.3 Effect of Water in the Backfill 709 14.8.4 Stability Reguirements 710 14.8.5 Soil Bearing Pressure Requirements 71 1 14.9 Proportioning and Design of Cantilever and Counterfort Walls 712 14.9.1 Position of Stem on Base Slab for Economical Design 712 14.9.2 Propotlioningand Design of Elements of Cantilever Walls 714 14.9.3 Proportioning and Design of Elements of a Countelfort Wall 715 Review Questions 745

~

~

~~

load

7nk

. .

16.3.8 shear wails (Flexural Wails) 788 16.3.9 lnfili frames 790 36310 Soft storey 791 '16.3.1 1 Performance limit states 792 16.4 Closure 792 Review Questions 792 References 793

17. Selected Special Topics

Problems 746 References 747

15. GOOD DETAILING AND CONSTRUCTION PRACTICES

16 3.4 Foundations 780 16.3.5 Flexural Members In Ductlie Frames 780 16.3.6 Columns and'frame Members Subiect to Bendinfl and Axial

XXI

749

15.1 lntroduction 749 15.1.I Se~iceabilityFailures 750 15.1.2 Reasons for Building Failures 751 15.1.3 Structural Integrity 751 3.2 Design and Detailing Practices 752 15.2.1 Reinforcement Layout 753 15.2.2 Design Drawings 754 15.2.3 Construction Details at Connections and special situations 754 15.2.4 Beam and Column Joints (Rigid Frame Joints) 761 15.2.5 Construction Joints 763 15.2.6 Bar Supports and Cover 764 15.2.7 Deflection Control 765 15.3 Materials and Construction Practices 765 15.4 Summary 767 Review Questions 768

Design for Shear b y Compression Field Theory 795 lntroduction 795 General Concepts 796 Stress-Strain Relationship for Diagonally Cracked Concrete 798 Analysis Based on Modified Compression Field Theoly 799 Simplified Design Procedure using Modified Compression Field Theory 804 CSA Code provisions for Shear Design by the Compression Field Theory 808 Combined Shear and Torsion 810 17.2 Design Using Strut-and-Tie Model 811

17.3 Fire Resistance 822 17.3.1 lntroduction 822 17.3.2 Factors which influence Fire Resistance Ratings of RC Assemblies 823 17.3.3 Code Requirements 825 Problems 826 Review Questions 826 References 827

References 769

16. SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANT DESIGN 771 16.1 introduction 771 16.2 lmportapce of Ductility i n Seismic Design 773 16.2.1 Measyres of Ductility 773 16.2.2 Energy Dissipation by Ductile Behaviour 775 16.2.3 Flexural Yielding in Frames and Walls 777 16.3 Major Design Considerations 778 16.3.1 General Design Objectives 778 16.3.2 Requirements of Stability and Stiffness 778 16.3.3 Materials 779

APPENDIX A: ANALYSIS AND DESIGN AIDS

829

Table A.l ANALYSIS AIDS (WSM) for Singly Reinforced Rectangular Beam Sections Vaiues of ~,,,lbd* (MPa) for given Vaiues of p, (a) M 20,M 25 Concrete Grades 830 (b) M 30,M 35 Concrete Grades 831 Table A.2 ANALYSIS AlDS (LSM) for Singly Reinforced Rectangular Beam Sections. ~ ~given a ) Values of p, Values of ~ , , ~ l b d * ( for

(a) M 20,M 25 Concrete Grades 832 (b) M 30,M 35 Concrete Grades 835

XXll

,

CONTENTS

,

Table A.3 DESIGN AIDS (LSM) for Singly Reinforced Rectangular Beam SectionsValues of p,for given Values of R - ~ , l b d ' ( ~ p a ) (a) M 20, M 25 Concrete Grades 839 (b) M 30, M 35 Concrete Grades 843 Table A.4 DESIGN AIDS (LSM) for Doubly Re~nforcedRectangular Beam Sections Values of p, and p,for glven Values of R - ~ ~ b (MPa) d ' for (a) Fe 415 Steel, M 20 Concrete 849 (b) Fe 415 Steel, M 25 Concrete 855 2

TableA.5 Areas (rnm ) of Reinforcing Bar Groups 861 Table A.6 Areas (rnm2/m)of Uniformly Spaced Bars 862

APPENDIX B: GENERAL DATA FOR DEAD LOADS AND LlVE 863 LOADS Table B.l Table 8.2 Table 8.3 Table 8.4

Index

DEAD LOADS - Unit Weights of Some MaterialsIComponents 864

LlVE LOADS on Floors 865 LlVE LOADS on Roofs 865 HORIZONTAL LlVE LOADS on Parapets/Baluslrades 865 1.1 INTRODUCTION

Traditionally, the study of reinforced concrete design begins directly with a chapter on materials, followed by chapters dealing with design. In this hook, a departure is made from that convention. It is desirable for the student to have first an overview of the world of reinforced concrete structures, before plunging into the finer details of the subject. Accordingly, this chapter gives a general introduction to reinforced concrete and its .?pplicatiom It also expIains the role of stnrctural design in reinforced concrete construction, and outlines the various structural systems that are commonly adopted in buildings. That concrete is a common structural material is, no doubt, well known. But, how common it is, and how much a part of our daily lives it plays, is perhaps not well known - orrather, not often realised. Structural concrete is used extensively in the construction of various kinds of buildings, stadia, auditoria, pavements, bridges, piers. breakwaters, berthing structures, dams, waterways, pipes, water tanks, swimming pools, cooling towers, bunkers and silos, chimneys, communication towers, nmnels, etc. It is the most commonly used consttuction material, consumed at a rate of approximately one ton for every living human being. "Man consumes no material except water in such tremendous quantities" (Ref. 1.1). Pictures of some typical examples of reinforced concrete structures are shown in Figs 1.1-1.5. Pcrhaps, some day in the future, the reader may be callcd upon to design similar (if not, more exciting) strnctures! The stndent will do well to bear this goal in mind.

2 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

STRUCTURES 3

-

Flg. 1.1 Ferrocement Boat "the first known example of reinforced concrete" is a boat, patented in 1848 by Joseph-Louis Lambof [Ref. : Ferrocement, National Academy of Sciences. Washington D.C., Feb. 19731: the boat shown here is a later version (1887) of the original design, presently preselved in the Brignoies Museum, France.

Fig. 1.2 A modern reinforced concrete multi-storeyed building - one of the tallest in New Delhi (102 rn) : Jawahar Vyapar Bhavan [Architects : Raj Rewal andKuldlp Singh, Project Consultants : Engineers India Limited]. Structural concept : joist floor supported on Vierendeei girders (arranged in a 'plug on' fashion), cantilevered from core Walls.

Fig. 1.3 The BahB'i House of ~ o r s i i p New , Delhi - aunique lotus-shaped reinforced concrete structure, with a complex shell geometry involving spheres, cylinders, torroids and cones [Architect : Fariburz Sahba, Structural Consultants : Flint & Neill, Contractor : Larsen & Toubro Lld,]

Fig. 1.4 C N Tower - a communications tower at Toronto, Canada, rising to a height of 550 m, making it the tallest reinforced concrete tower in the world. (The picture also shows an eievator car wllich travels vertically aiong the shaft of the tower).

4 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

STRUCTURES 5

1.2.2 Reinforced C o n c r e t e

Fig. 1.5 A reinforced concrete bow-string girder bridge spanning across the Bharathapuzha River at Kultippuram, Keraia

1.2 PLAIN AND REINFORCED CONCRETE 1.2.1 Plain C o n c r e t e

Concrete is defined [Ref. 1.21 as any solid mass made by the use of a cementing medium; the ingredients generally comprise sand, gravel, cement and water. That the mixing togcther of such disparate and discrete materials can result in a solid mass (of any desired shape), with well-delincd properties, is a wonder in itself. Concrete has . been in use as a building material for more than a hundred and fifty years. Its success and popularity may bc largely attributed to (1) durability under hostile environmcnts (including resistance to water), (2) ease with whicll it can be cast into a variety of shapes and sizes, and (3) its relative economy and easy availability. The n~ain strength of concrete lies in its compression-hearing ability, which surpasses that of traditional materials like brick and stone masonry. Advances in conmete technology, during the past Four decades in particular, have now made it possible to produce a wide range of concrete grades, varying in mass dcnsity (1200-2500 kg/m3) and compressive strength (I0 -100 MPa). Concrete is remarkably strong in conrpression, hut it is equally rcmarknbly weak in tension! [Fig. 1.6(a)]. Its tensile 'strength' is approximately one-tenth of its compressive 'strength'. Hence, the use of plain concrete as a structm'al matcrial is limited Lo situations whcre significant tensile stresses and strains do not dcvelop, as in hollow (or solid) block wall construction, small pedestals .and 'mass concrete' applications (in dams, ctc.).

Concrete would not have gained its present status as a principal building material, but for the invention of reinforced concrete, which is concrete with steel bars embedded in it. The idea of reinforcing concrete with steel has resulted in a new composite material, having the ability to resist significant tensile stresses, which was hitherto impossible. Thus, the construction of load-bearing flexural members, such as beams and slabs, became viable with this new material. The steel b a n (embedded in the tension zone of the concrete) compensate for the concrete's inability to resist tension, effectively taking up all the tension, without separating from the concrete [Fig. 6 ( b The bond between steel and the surrounding concrete ensures strain compatibility, i.e., the strain at any point in the stee1.i~equal to that in the adjoining concrete. Moreovei, the reinforcing steel imparts ductility to a material that is otherwise brittle. In practical terms, fhis implies that if a properly reinforced beam were to fail in tension, then such a failure would, fortunately, he preceded by large cleflections caused by the yielding of steel, thereby giving ample warning of the impending collapse [Fig.l.6(c)J. Tensile stresses occur either directly, as in direct tension or flexural tension, or indirectly, as in shear, which causes tension along diagonal planes ('diagonal tension'). Temperature and shrinkage effects may also induce tensile stresses. In all such cases, reinforcing steel is essential, and should he appropriately located, in a direction that cuts across the principal tensile planes (i.e., across potential tensile cracks). If insufficient steel is provided, cracks would develop and propagate, and could possibly lead to failure. Being mt~chstronger than concrete in compression as well, reinforcing steel can also suonlement concrete in bearine comoressive forces. as in columns orovided with longitudinal bars. These bars need to he confined by transverse stccl ties [Fig. 1.6(d)], in ordcr to maintain their positions and to their lateral buckling. The lateral ties also serve to confine the cmcrcte. therehv enhancinz - its comorcssion load-bearing capacity. As a result of extensive research on reinforced concrete over the past several decades in various countries, a stage has reached where it is now possible to predict the elastic and inelastic behaviour of this composite material with some confidence. No doubt, there exists some uncertainty in the prediction, but this is largely attributable to the variability in the strength of in-situ concrete (which, unlike steel, is not manufactured under closely cont~olledconditions). There are several factors which lead to this variability, some of which pertain to material properties (primarily of the aggregates), while others pertain to the actual making of concrete at site (mixing, placing, compacting and curing). This uncertainty can be taken care of, by providing an appropriate factor of safety in the design process. [The topic of structural safety in design is discussed in detail in Chapter 31. The development of reliable design and construction techniques has cnabled the construction of a wide.variety of reinfoxed concrete structures all over the world: building frames (columns andbeams), floor and roof slabs, foundations, bridge decks and picrs, rztaining walls, grandstands, water tanks, pipes, chimneys, bunkers and silos, folded plates and shells, etc.

..

- .

REINFORCED

6 REINFORCED CONCRETE DESIGN

(a) Plain concrete beam

cracks and falls In flexural tension under a small load

(b) Reinforced concrete

Steel bUS embedded

A

beam supports loads with acceptably low defo;mations

\ halrllne crack

(not perceptible)

(0)

Ductlle mode of fallure under heavy loads

steel bars undergo yielding longitudinal reinforcement tunder carn~ression)

CONCRETE

STRUCTURES 7

It is worth noting that, although these reinforced concrete structures appear to be completely diffemnt from one another, the actual principles underlying their. design are the same. In the. chapters to follow, tlie focus will be on tliese.fundamental principlds.

Prestressed Concrete: An inlroduction to reinforc'dd concrete will notbe complete without amention of prextressed concrete, which is another ingenious invention that developed side-by-side with reinforced concrete. Prestressed concrete is highstrength concrete with high tensile steel wires embedded and tensioned, prior to the application of external loads. By. this, the concretecan be pre-compressed to such a degree that, after the structure is loaded, thek is practically no resultant tension developed in the beam. Prestressed concrete Khds application in situations where long spans are encountered (as in bridges), o r where cracks (even hairline) in concrete are not permitted (as in pressure vessels, pipes and water tanks), or where fatigue loading is encountered (as in railtrack sleepers), etc. Fibre-Reinforced Concrete and Ferrocement: Rccent developments in concrete composites havc resulted in several new products that aim to improve the teiisile strength of concrete, and to impart ductility. Among thcse, fibre-reinforced concrete and fermxmncnt constilutc important developn~ents. In the forniel-, steel or glass fibres are incorporated in concrete at the time oiiniixing; in thc latter, thin sections are fanned by embedding multiple layers of steel wire mesh in cement mortar. Although ferrocenient has gained popularity only in rccent years, it reprreseuts one of the earliest applications of reiuforced concrete to be experiniented with [Fig. 1.11. This book is concerned with reinforced concrete; hence, no further discussion on other concrete composites will bc made.

1.3 OBJECTIVES O F STRUCTURAL DESIGN The design of a structure must salisfy thrce basic requirements: 1) Stability to prevenl ovcrluming, sliding or buckling of the structure, or parts of it,

under the actioii of loads,

2) Strength to resist safely the stresses induced by the loads in the various structural members; and

3) Serviceability to ensum satisfactory performance under service load conditions

-

which implies providing adequate stg'ress and reinforcements to contain deflections, cmack-widths and vibrations within acceptable limits, and also provldi;ig irrrper.rneabilily aud rlrrrabilitj (including corrosion-resistance), etc. There are two other considerations that a sensible designer ought . . to benr in mind, viz., economy and nesthetics. One can always dcsign a massive structure, which has more-than-adeauate stabilitv, strenelh - and serviceabilitv, but the ensuine cost of the structure may be exorbitant, and the end product, far irom aesthetic. In the words of Fclix Candela [Ref. 1.31, the designer of a rcmarkably wide range of reinforced concrete shell structures,

-

Fig. 1.6 Contribution of steel bars in reinforced concrete

8 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

STRUCTURES 9

During the construction phase, some redesign may also be required - in the even1 of unforeseen contingencies, such as complications in foundations, non-availability of specilied materials, ctc. It is indeed a challenge, and a responsibility, for the structural designer to design a structorc that is not only appropriate for thc architecture, hnt also strikes thc right balance betwccn safcty and economy [Ref. 1.41.

1.4 REINFORCED CONCRETE CONSTRUCTION Reinforced concrete construction is hot the outcome of structural design alone. It is a collab?rative venture involving the client, the architect, the structural engineer, the construction engineedproject manager and. the contractor. Other specialists may also have to be consulted, with regard to soil invcstigatiou, water sopply, sanitation, fire protection, transportation, hcating, ventilation, air-conditioning, acoustics, electrical services, etc. Typically, a construction project involves three phases viz. planning, design (including analysis) and construction. 1. Planning Phase: It is the job of the architect/planncr to conceive and plan the architectural layout of the building, to suit the functional requirements of the client, with due regard to aesthetic, cnvironrnental and economic considerations. Structural feasibility is also an important consideration, and for this the structoral designer has to be consulted. 2. ~ e s i g nPhase: Once the preliminary plans have been approved, the actual details of the project have to be worked out (on paper) by the various consultants. In the case of the structural engineedconwltant, the tasks involved are (i) selection of the most appropriate structural system and initial proportioning of members, (ii)esdmation of loads on the structure, (iii) structural analysis for the determination of the stress resultants (member forces) and displacements induced by various load combinations, (iv) strnetural design of the actual proportions (member sizes, reinforcement details) and grades of materials required for safety and serviceability under the calculated member forces, and (v) submission of working drawings that are detailed enough to be stamped 'good for construction'. 3. Construction Phase: The plans and designs conceived on paper get translated into concrete (!) reality. A structure may be well-planned and well-designed, but it also has to be well-built, for, the proof of the pudding lies in the eating. And for this, the responsibility lies not only with the contractor who is entrllsted with the execution, but also with the construction engineers who undenake supervision on behalf of the consultants. The work calls for proper management of various resources, viz. manpower, materials, machinery, money and time. It also requires familiarity with various construction techniques and specifications. In particular, expertise in concrete technology is essential, to ensure the proper mixing, handling, placing, compaction and curing of concrete. Management of contracts and following proper procedures, systems and documentation are also important aspects of the construction phase, especially in public works, however these are beyond the scope of this book.

1.5 STRUCTURAL SYSTEMS Any structure is made up of structural elements (load-canying, such as beams and columns) and non-srructural elements (such as partitions: false ceilings, doors). The structural elements, put together, constitute the 'structural system'. Its function is to resist effcctively the action of gravitational and envil.onmentai loads, and to transmit the resulting forces to the supporting ground, without significantly -disturbing the geometry, integrity and serviceability of the structure. Most of the structssal elements may be considered, from the viewpoint of simplified analysis, as one-dimensional (skeletal) elements (such as beams, columns, arches, truss elkinents) or two-dirnerrsional elements (such as slabs, plates and shells). A lew stl.ucturaI dements (such as shell-edge beam junctions, perforated shea~walls) may more rigorous analysis. . require . Consider, for example, a reinforced concrete overhead water tank structure [Fie. . -1.71. . The structural svstcm essentiallv comorises three suhsvstems. viz. the tank. the staging and the foundation, which are distinct from one another in the sense that they are generally designed, as well as constructed, in separate stages. The tank, in this example, is made up of a dome-shaped shell roof, a cylindrical side-wall (with stiffening ring beams at top and bottom), a flat circular base slab, and a main ring beam, which is snpported by the columns of the staging. The staging comprises a three-dimensional framework of beams and columns, which are 'fixed' to the foundation. The foundation is a 'raft', comprising a slab in the shape of an annular rine. stiffened hv a rine beam on ton.. and restine on firm soil below. The loads actineu on the structure are due to dead loads (due to self-weight), live loads (due to water in the tank, maintenance on the roof), wind loads (acting on the exposed surface areas of the tank and staeinr). -.. and seismic loads (due to earthauake induced rround excitation). The effect of the loads acting on ;he tank are transmitted to the itaging through the main ring beam; the effect of the loads on the staging are, in turn, transmitted to the foundation, and ultimately, to the ground below.

-

v

.

-

-

1.6 REINFORCED CONCRETE BUILDINGS The most common reinforced concrete construction is the building (planned for residential, institutional or commercial use). It is therefore instructive to look at its structural system and its load transtnlssion mechanism in some detail. As the height of the building increases, lateral loads (due to wind and earthquake) make their presence felt increasingly; in fact, in very tall buildings, the choice of a structural system is dictated primarily by its relative economy in effectively resisting lateral loads (rather than gravity loads). For convenience, we may separate the structural system into two load transmission mechanisms, viz, gravity load resisting and lateral load resisting, although, in effect, these two systems are complementary and interactive. As an integratcd system, the structure runsfresist and transmit all the effects of gravity loads and lateral loads acting on it to the foundation and the ground below.

10 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

Moreover, although the building is a three-dimensional structure,-it is usually conceived, analysed and designed as an assemblage of two-dimensional (planar) subsystems lying p~inmarilyio the horizontal and vertical planes (e.g., floors, mof, walls, plane frames, etc.), as indicated in Fig. 1.8. This division into a horizontal (floor) system and a vertical (framing) system is pmlicularly convenient in studying the load resisting mechanislns in a building.

side wall

GRAVIN LOADS

STRUCTURES 11

TANK

1.6.1 Floorsystems

(c) section.through tank STAGING

The (horizontal) floor system resists the gravity loads (dead loads and live loads) acting on it and transmits these to the vertical framing system. In this process, the floor system is subjected p~imadlyto flexure and transverse shear, whereas the vertical frame elen~entsare generally subjected to axial compression, often conpled -with flexure and shear [Fig. 1.8aI. The floor also serves as a horizontal diaphragm connecting together and stiffening the various vertical frame elemnents. Under the action oflater,d loads, the floor diaphragm behaves rigidly (owing to its high in-plane flexural stiffness), and effectively distributes the lateral load cffects to the various vertical frame elenlents and shear walls [Fig. 1.8bl. In cast-in-situ rcinforced concrete construction, the floor system usually co~lsistsof one of the following: Wall-SupportedSlab System

'-

(a) elevation

FOUNDATION

soil pressures 2 (d) section fhrougll foundation

(b) plan of foundation

Fig. 1.7 Structural system of an elevated water tank

In this system, the floor slabs, generally 100-200 nun thick with spans ranging from 3 m to 7.5 m, are suppofled on load-bearing walls (masonry). This system is mainly adopted in low-rise buildings. The slab panels are usually rectangular in shape, and can be supported inn number of ways. When tlie slab is supported only on two opposite sides [Fig. 1,9(a)], the slab bends in one direction only; hence, it is called a om-way slab. When the slab is supported on all four sides, and the plan dimensions of length and breadth are comparable to each other [Fig. 1.9(c)], the slab bends in two directjmrs (along the length and along the breadth); hence, it is called a two-way slab. However, if the plan is a long rectangle (length greater than about twice the width), the bending along the longitudinal direction is ~~egligible ill, colnparison with that along the transverse (short-span) direction, and the resulting slab action is effectively one-way [Fig. 1.9@)]. If the wall extends above the floor level [Fig. 1.9(d)], the slab is no more simply supported; tlie partial fixity at the snpport introduces hogging moments in the slab. Furthermore, twisting moments are also introduced at the comers that are restrained (not free to lift up) - as established by the classical theory of plates. Generally, slabs are cast i n panels that are continuous over sevel-a1wall supports, and are called one-way cor~rLtuous[Fig. 1.9(e)] or nvo-wrry continuous slabs, depending on whether the bcnding is prcdorninantly along one direction or two directions. Hogging monrcnts are induced in the slab in the region adjacent to thc continuous support.

12 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

column 7

:I I

STRUCTURES 13

rbeam

all

PLAN

load-bearing wall

wo-way slab (d) hogging moments near end

support (partially fixed) (a) vertical load transmlsslon

rigid connection

.,

(€4 continuous slab

FLOOR SYSTEM as

horizontal diaphragm

Flg. 1.9 Wali-supported slab systems

Beam-Supported Slab System

cladding

\'r/ (b) lateral load transmission

Flg. 1.8 Load transmission mechanisms

Thissystem is similar to the wall-supported slab system, except that the floor slabs are supportedon beams (instead of walls). The beams are cast monolithically with the slabs in a grid pattern [Fig. l.lO(a)], with spans ranging from 3 m to 7.5 m. This system is commonly adopted in high-rise building construction, and also in low-rise f,nmed structures. The gravity loads acting on the dabs are transmitted to the colun~nsthrough the network of beams The beams which are directly connected to the columns (forming the vertical frames) are called primaiy beams (or girders); whereas, the beams which are supported, not by columns, but by other (primary) beams, are called secondary b e a m [Figs l.lO(b),(c)]. If the beams are very stiff, the beam deflections are negligible, and the slab supports become relatively unyielding, similar to wall supports; the action may be either two-way or one-way [Fig. l.lO(b),(c)l, depending on the panel dimensions. However, if the beams are relatively flexible, the beam deflections are no longer negligible and will influence the slab behaviour. When a large number of two-way secondary beams are involved (typically in a 'grid floor' with a large column-free space) [Fig. l.lO(d)], the slabs do not really 'rest' on the beams; the slab-beam system as a whole acts integrally in supporting the gravity loads.

REINFORCED CONCRETE STRUCTURES 15 14 REINFORCED CONCRETE DESIGN

Ribbed Slab System This is a special type of 'grid floor' slab-beam system, in which the 'slab', called topping, is very thin (50-100 nun) and the 'beams', called ribs, are very slender and closely spaced (less than 1.5 m apart). The ribs have a thic!aess of not less than 65 mm and a depth that is three-to-four times the thickness. The ribs may be designed in one-way or two-way patterns [Fig. I.ll(a),(b)], and are generauy cast-in-situ, although precast cons~mctionis also possible.

I

I

column

ribs

u

(a) beam-suppolled slab

<

PLAN

(view from below)

/-- primary beam

column

primary b e a m 1 I

ribs

< SECTION (enlarged)

'AA'

primary beams

beam

Fig. I.ll(a) One-way ribbed slab system (b) two-way system

(c) one-way system

Two-way ribbed slabs are sometimes called wajle slabs. Along the outer edges, the ribbed slab system is generally supported on stiff edge beams or walls. In wallsupported systems, the thickness of the rib resting on the wall is usually increased to match the wall thickness for improved bearing. Waffle slabs, used in large-span consuuction, may lest directly on columns; in this case, the slab is made solid in the neighbourhood of the columm.

Flat Plate System

(d) grid beam-suppolled slab

Fig. 1.10 Beam-supported slab systems

Here, the floor slab is supported directly on the columns, without the presence of stiffening beams, except at the periphery [Fig. 1,121. It has a uniform thickness of about 125-250 mm for spans of 4.5-6 m. Its load carrying capacity is restricted by the limited shear strength and hogging moment capacity at the column supports. Because it is relatively thin and has a flat under-surface, it is called aflat plate, and certainly has much architectural appeal. It is used in the developed countries at locations (in apartments and hotels) where floor loads are low, spans are not large, and plane soffits serve as ceilings. However, it is yet to gain popularity in India - perhaps, because it is too daing a concept?

REINFORCED CONCRETE STRUCTURES

17

Flat Slab System

< (view from below) PLAN

This i s a more acceptable concept to many designers [Fig. 1.131. It is adopted in some office buildings. The flat slabs are plates that are stiffened near the column supports b y means of 'drop panels' andlor 'column capitals' (which are generally concealed under 'drop ceilings'). Compared to the flat plate system, the flat slab system is suitable for highe~lokdsand larger spans, because of its enhanced capacity in resisting shear and hogging moments near the supports. The slab thickness varies from 125 nun to 300 mm for spans of 4-9 m. Among the various floor systems, the flat slab system is the one with the highest dead load per unit area.

Fig. l.ll(b) Two-way ribbed (waffle)slab system Fig. 1.13 Flat slab system 1.6.2 Vertlcal Framlng System

As mentioned earlier, the vertical framing system resists the gravity ioads and lateral loads from the floor system and transmits these effects to the foundation and ground below. The framingsystem is made up of a three-dimensional framework of beams and columns. For convenience, we may divide the framework into separate plane frames in the transverse and longitudinal directions of the building. In cast-in-situ reinforced concrete construction, the vertical framing system usually comprises the following:

columns

Flg. 1.12 Flat plate system

These are skeletal structural elements, whose cross-sectional shapes may be rectangular, square, circular, Lshaped, etc. - often as specified by the architect. The size of the column section is dictated, from a structural viewpoint, by its height and the loads acting on it - which, in turn, depend on the type of floor system, spacing of columns, number of storeys, etc. The column is generally designed to resist axial compression combined with (biaxial) bending moments that are induced

18 REINFORCED CONCRETE DESIGN

by 'frame action' under gravity and lateral loads. These load effects ate more pronounced in the lower storeys of tall buildings; hence, high strength concrete (up to 50 MPa) with high reinforcement area (up to 6 percent of the concrete area) is frequently adopted in such cases, to minimise the column size. In some situations, the column height - between floor slabs mav be excessive (more than one storev. heightk - . in such cases, it is structurally desirable to reduce the unsupported length of the column by providing appropriate tie beam; otherwise, the columns should be propedy designed as slender columns. Walls These are vertical elements, made of masonry or reinforced concrete. They are called bearing walls if their main structural function is to support gravity loads, and are referred to as shear walls if they are mainly required to resist lateral loads due to wind and earthquake. The thickness of reinforced concrete bearing walls varies from 125 mm to 200 mm: however, shear walls may be considerably thicker in thc lower storeys of tall buildings. The walls around the lift cores of a building often serve as shear walls.

REINFORCED CONCRETE

STRUCTURES 19

Suspenders These are vertical elements used to suspend floor systems such as the cantilevered upper storeys of a multi-storeyed building from a central reinforced concrete core [Fig. 1.151. Structural stccl is often found to be better suitcd for dse as suspenders (also called hangers), because the force to be resisted is direct tension; moreover, steel hangers takc up vcry little of the floor space. The loads from the suspenders may be transmitted to thc reinforced concrete core by means of large cantilevered beams, cross-braccd trusscs or Vierendeel girders [also refer Fig. 1.21.

,Vierendeel girder

'L

suspenders

Transfer Girders In some buildings, the architectural planning is such that large column-free spaces are required in the lower floors - for banquetkonvention halls (in hotels), lobbies, parking areas, etc. In such cases, the vertica! load-bearing elements (columns, bearing walls) of the . ' upper floors are not allowed to continue downwards, through the lower floors, to the foundations below. This problem can be resolved by providing a very heavy beam, called transfer girder, whose depth may extend over one full storey [Fig. 1.141. The upper-storey columns terminate above the transfer girder, and transmit their loads, through the beam action of the girder, to the main columns that support the girder from below.

cantilevered 1100

central care wall

Fig. 1.15 Use of suspenders

It may bc noted that the vertical elements in the bow-string girder of Fig. 1.5 also act as suspenders, t r a n s ~ ~ t t i nthe g loads of the bridge deck to the arches spanning between the piers.

1.6.3 Lateral Load Resisting S y s t e m s As mentioned earlier, the horizontal and vertical sub-systems of a structural system interact and jointly resist both gravity loads and lateral loads. Lateral load effects (due to wind and earthquake) predominate in tall buildings, and govern the selection of the structural system. Lateral load msisting systems of reinforced concrcte buildings generally consist of one of the following: Frames Flg. 1.14 Use of transfer girder

These are generally conq~oscdof columns and beams [Fig. I.X(b) and 1.16(a)l. Their ability to resist lateral loads is entirely due to the rigidities of thc bcam-column connections and the momeot-resisting capacities o l [he individual mcmbeu They are often (albeit mistakenly) called 'rigid frames', because the ends of the various members framing into a joint are 'rigidly' connected in such a way as to ensure that

20

REINFORCED

CONCRETE DESIGN

REINFORCED CONCRETE

they all undergo the same rotation under the action of loads. In the case of the 'flat plate' or 'flat slab' system, a certain width of the slab, near the column and along the column line, takes the place of the beam in 'frame action'. Frames are used as the sole lateral load resisting system in buildings with up to 15 to 20 storeys [Fig. l.l6(e)].

(a) rigid frame

(b) shear wall

(c) shear wall - frame

action

action

interaction

STRUCTURES 21

Shear Walls These are solid walls, which usually extend over the full height of the building. They are commonly located at the liftlstaircase core regions. Shear walls are also frequently placed along thc transverse direction of a building, either as exterior (facade) walls or as interior walls. The walls are very stiff, having considerable depth in the direction of lateral loads [Fig. 1.16(b)]; they resist loads by bending likc vertical cantilevers, fixed at the base. The various walls and co-existing frames in a building are linked at the different floor levels by means of the floor system, which distributes.the lateral loads to these different systems approptiately. The interaction between the shear walls and the franles is structurally ady,atltageous i n that the walls restrain the frame deformations in the lowi+store~s,.wh~eihe frimis iesUain the wall deformations in the upper storeys [Fig. l.lk(c)]: Frame-shear wall systems are generally considered i n buildings up to about 40 storeys, as indicated in Fig. 1.16(e) [Ref. 1.51.

Tubes (d) tube

u action

These are systems in which closely-spaced columns are located along the periphery of a building. Deep spandrel beams, located on the exterior surface of the building, interconnect these columns. The entire system behaves like a perforated box or framed tube with a high flexural rigidity against lateral loads [Fig. 1.16(d)]. When the (outer) framed tube is combined with an 'inner tube' (or a central shear core), the system is called a tube-in-tube. When the sectional plan of the building comprises several perforated tubular cells, the system is called a bundled tube or 'mnulti-cell framed tube'. Tubular systems are effective up to 80 storeys, as indicated in Fig. 1.16(e). Widely adopted in the big cities of developed countries, these skyscraping systems are on the verge of making an appearance in the metros of India.

BUNDLW TUBE

1.7 STRUCTURAL ANALYSIS AND DESIGN

It is convenient to separate the work of a stmctural designer into analysis and design, although a rigid separation is neither possible nor desirable. When a student undergoes separate courses on structural analysis and structural design, it is essential that he realises the nature of their mutual relationship. The purpose of analysis is n determine the stress r e s u ~ t a ~ a & ~ e k... n y i ~ ~ the various ._ members of a structure . ! ? . The ......^,,*I__" _.-_--__.) --.- under a ~ ~ ~ % t ~ s , , y~.a,wi). purpose oThesign is to pro= adequate member sizes, reinforceme2 and connection details, so as to enable the structure to d I X d i W l f ~ l &EiIcufated y Ioad'efEts. In order to perform analysis, the proportions of the various structural elements should be known in advance; for this, a preliminary design is generally required. Thus, in nracHce. analvsis and desien are interactive Drocesses

----".-..-.-*

--

,-,.,-A

(e) comparison ofvarioussystems

Fig. 1.16 Lateral load resisting systems

22 REINFORCED CONCRETE DESIGN

REINFORCED CONCRETE

STRUCTURES 23

Furthermore, the various methods of analysis of structures [Ref. 1.6-1.91 clearly lie outside the scope of this book. However, some approximations in analysis, as permitted by design codes, are discussed in some of the chapters to follow.

This code shall henceforth be referred to as 'the Code' in the chapters to follow. pefere~aeshave also been made to other national codcs, such as ACI 318, B S 8110, CSA CAN3-A23.3 and Eurocode, wherever relevant.

Exposure to Construction Practices

1.8.3 Loading Standards

In rernforced concrete structures, construction practices are as Important as the design. Indeed, for a correct understanding of design as well as the Code provisions, some exposure to concrete laboratory work and to actual reinforced concrete construction work m the field is required.

The loads to be considered for struclural design are specified in the following loading standards:

1.8 DESIGN CODES AND HANDBOOKS 1.8.1 Purpose of Codes National building codes have been formulated in different countries to lay down guidelines for the design and construction of structures. The codes have evolved from the collective wisdom of expert structural engineers, gained over the years. These codes are periodically revised to bring them in Line with current research, and often, current trends. The codes serve at least four distinct functions. Firstly, they ensure adequate structural safety, by specifying certain essential mini~numrequirements for design. Secondly, theyrender thc task of the designer relatively simple; often, the results of sophisticated analyses arc made available in the form of a simple formula or chart. Thirdly, the codes ensure a measure of consistency among different designers. Finally, they have some legal validity, in that they protect the structural designer from any liability due to structural failures that are caused by inadequate supervision and/or faulty material and construction.

1.8.2 Basic Code for Design The design procedures, described in this book, conform to the following Indian code for reinforced concrete design, published by the Bureau of Indian Standards, New Delhi: IS 456 : 2000 - Plain and reinforced concrete - Code of practice (fourth revision)

:

IS 875 (Parts 1-5) : 1987 - Code of practice for design loads (other than earthquake) for buildings and structures (second revision) P a r t 1 : Dead loads P a r t 2 : Imposed (live) loads P a r t 3 : Wind loads Part 4 : Snow loads P a r t 5 : Special loads and load combinations IS 1893 : 2002 - Criteria for earthquake resistant design of structures (fourth revision).

1.8.4 Design Handbooks The Eureau of Indian Standards has also published the following handbooks, which serve as useful supplcnlents to the 1978 version of the Code. Although the handbooks need to be updated to bling them in line with the recently revised (2000 version) of the Code, many of the provisions continue to be valid (especially with regard to structural design provisions). S P 16 : 1980 -Design Aids (for Reinforced C o l l c ~ ~ tto e )IS 456 : 1978 S P 24 : 1983 -Explanatory Handbook on IS 456 : 1978 S P 34 : 1987 - Handbook on Concrete Reinforcement and Detailing S P 2 3 : 1982 -Design of Concrete Mixes

1.8.5 Other Related Codes There are several other codes that the designer may need to refer to. The codes dealing with material specifications and testing are listed at the end of Chapter 2. Chapter 16 of this book deals with special design provisions related to earthquakeresistant design of reinforced concrete structures. The code related to this topic is: IS 13920 : 1993 - Ductile detailing of reinforced concrete suucturcs subjected to seismic forces. Other codes dealing with the design of special structures, such as liquid-retaining structures, bridges, folded plates and shells, chimneys, bunkers and silos, are not covered in this book, the scope of which is limited to basic Ceinforced concrete design.

..

i:

i '

REVIEW QUESTIONS 1.1 What reasons do you ascribe to concrete gaining the status of the most widely used construction material? 1.2 The occurrence offlex~trultension in reinforced concrete is well known. Cite practical examples where tension occurs in other forms in reinforced concrete. 1.3 What is the role of transverse steel ties [Fig. 1.6(d)] in reinforced concrete co1umls? 1.4 A reinforced concrete canopy slab, designed as a cantilever, is under construction. Prior to the lemoval of the fonnwork, doubts are expressed about the safety of the stmcture. It is proposed to prop up the free edge of the cantilever with a beam supported on pillars. Comment on this proposal. 1.5 What are the main objectives of structural design? 1.6 List the steps involved in the process of structural design. 1.7 Distinguish between structural design and structural analysis. 1.8 Consider a typical reinforced concrete building in your institutiotl. Identify the various strllctllral elements in the structural systcm of the building, and briefly explain how the loads are transmitted to the supporting ground. 1.9 Consider a symmetrical portal frame ABCD with the columns (AB and CD) 4 m high, fixed at the base points A and D. The beam BC has a span of 6 tn and supports a uniformly distributed load of 100 liN. From str~lcturalanalysis, it is found that at each fixed base support, the reactions developed are 50ldrT (vertical), 30 kN (horizontal) and 40 ldrT m (moment). With the help of keebody, bending moment, shcar force and axial force diagrams, determine the stress resultants in the design of the beam BC and the column AB (orCD). 1.10 Enumerate the various types of gravity load bearing systems and lateral load resisting systcms used in reinforced concrete buildings. REFERENCES 1.1 Mehta, P.K. and Monteiro, P.I.M., Conorre: Microstr.uctlrre, Properries and Muteriuls, It~dianedition, Indian Concrete Institute, Chennai, 1997. 1.2 Neville, A.M. and Brooks, J.J., Concrete Technology, ELBS edition, Longman, London, 1990. 1.3 Faber, C., Candela, The ShellBuilder, Architectural Press, London, 1960. 1.4 Salvadori, M. and Heller, M., Structure in Architecture, Prentice-Hall International 1986 . ~....., ~ -1.5 Fintel. M., Handbook of Concrete EngLteeritrg, Second edition, (Van Nostra~ld Co., New York), C.B.S. Publishem and Distributors, Delhi, 1986. 1.6 Wang, C.K., Intermediate Structural Analysis, McGraw-Hill International edition, 1983. 1.7 Weaver, W. and Gere, J.M., Matrix Atralysis of F~'nnredStructures, Second edition, Van Nostrand Co., New York, 1980. 1.8 Clough, R.W. and Penzien, J., Dynamics of Sh-mmrcs, Second edition, McGraw-Hill International edition, 1993. 1.9 Talanath, B.S., SttucnmralAnalysis mrdDerigr~of TnllBuildings, McGraw-Hill International edition, 1988.

In order to learn to design reinforced concrete structures, it is desirable to begin with an understanding of the basic materials, viz. concrete (inclnding its ingredients) and reinforcing steel. Accordingly, this chapter describes briefly some of the important properties of these basic materials. Much of this chapter is devoted to concrete rather than steel, because the designer (as well as the builder) needs to know more about concrete, which, unlike steel, is not manufactured in factories under controlled conditions. Concrete is generally prepared at the site itself, although precast concrete is also used in some cases. 2.1 .I Concrete Technology ~

The making of 'good' concrete is decidedly not an easy job. This is clear from the all-too-common >ad' concrete. Both good and bad concrete can be prepared from exactly the same constituents: cement, aggregate, water (and, sometimes, admixtures). ~t is the mixproportions, the 'know-how' and the 'do-how' that makes the difference. Good concrete is one that has the desired qualities of strength, impermeability, durability, etc., in the hardened state. To achieve this, the concrete has to be 'satisfactory' in the fresh state (which includes mixing, handling, placing, compacting and curing). Broadly, this means that the mix must be of the right proportions, and must be cohesive enough to be transported and placed without segregarion by the means available, and its consistency must be such that it is workable and can be compacted by the means that are actually available forthe job. A cotnpetent concrete technologist will be able to get a fair idea of the nature and properties of hardened concrete by observation and a few simple tests on the fresh concrete. If found unsatisfactory, suitable remedial measures can and should be adopted without having to wait until the concrete hardens, by which time it is to.0 late to effect corrections

26 REINFORCED CONCRETE DESIGN

'Concrete technology' is a complete subject in itself, and the reader is advised to consult standard textbooks on the subject [Ref. 2.1, 2.2, 2.31 for a detailed study. In the following sections, some salient features of the making of concrete (covering both ingredients and process) are discussed, followed by a detailed description of the properties of hardened concrete and reinforcing steel. 2.2 CEMENT

Cement may bedescribed as a material with adhesive and cohesive properties that make it capable of bondingmine~alfragments, ('aggregates') into a compact whole [Ref. 2.11. In this process, it imparts strengih and dvmbiliiy to thehardened mass called 5,oncrete. The cements used in the making of concrete are called hydraulic cenrents - so named, because they have the property of reacting chenlically with 'water in an exothermic (heat-generating) process called hydration that results in water-resistant products'. The products of hydration form a viscous cement paste, which coats the aggregate surfaces and fills some of the void spaces between the aggregate pieces. The cement paste loses consistency ('stiffens') on account of gradual loss of 'free water', adsorption and evaporation, and subsequently 'sets', t~ansformingthe mixture into a solid mass. If the consistency of the cement paste is either excessively 'harsh' or excessively 'wet', there is a danger of segregation, i.e., the aggregate tends to separate out of the mix; this will adversely affect the quality of the hardened concrete and result in a 'honeycomb' appearance. The freshly set cement paste gains strength with time ('hardens'), on account of progressive filling of the void spaces in the paste with the reaction products, also resulting in a decrease in porosity and permeability.

2.2.1 Portland Cements

The most common type of hydraulic cement used in the manufacture of concrete is known as Portland cement, which is available in various forms. Portland cement was first patented in England in 1824, and was so named because its grey colour resembled a limestone (quarried in Dorset) called 'Portland stone'. Portland cement is made by burning together, to about 1400°C, an intimate mixture (in the form of a slurry) of limestone (or chalk) with alumina-, silica- and iron oxidebearing materials (such as clay or shale), and grinding the resulting 'cli&er' into a fine powder, after cooling and adding a little gypsum. The cement contains four

'

Cements derived from calcination of gypsum or limestone are 'non-hydraulic' because their products of hydration are not resistant to water; however, the addition of pozzolaiic materials can render gypsum and lime cements 'hydraulic' [Ref. 2.21.

BASIC

MATERIAL

PROPERTIES 27

major cotnpounds, viz., tricalcium silicate (C,S): dicalcium silicate (CZS), tricalcium aluminate (C3A) and tetracalcium aluminoferrite (CdAF). By altering the relative proportions of these major compounds, and including appropriate additives, different types of Portland cement, with different properties, can be made. For instance, increased proportions of C3S and C,A contribute to high early strength; on the contriry, an increased proportion of C2S retards the early development of strength (and generates less heat of hydration), but enhances ultimatc strength [Ref. 2.21. Adjusting the fineness of cctncnt can also control these propertics. The use of any one of the following types of Portland cement is permitted by the Code (IS 456 : 2000):

Ordinary Portland Cement (OPC) - presently available in three different 'grades' (denoting compressive strength), viz. C33, C43 and C53, conforming to IS 269 : 1989, IS 8112 : 1989 and IS 12269 : 1987 respectively. The numbers 33,43 and 53 cotrespond to the 28-day (charactc~istic')cotnpressive strengths of cemcnt, as obtained from standard tests on cement-sand mortar specimens. These are most commonly used in general concrete construction, where there is no special durability requirement (such as exposurc to 'sulphate attack'). Rapid Hardening Portland Cement (RHPC) -conforming to IS 8041 : 1990, is similar to OPC, exccpt that it has more C,S and less C,S, and it is ground more finely. It is used in situations whe~ca rapid development of strength is desired (e.g., when formwork is to be removed carly for reuse). Portland Slag Cement (PSC) - conforming to IS 455 : 1989, is made by intergrinding Portland cement clinker and granulated blast furnace slag (which is a waste product in the manufacture of pig iron). It has fairly high sulphate resistance, rendering it suitable for use in environments exposed to sulphates (in the soil or in ground water). Portland Pozzolana Cements (PPC) - flyash based or calcined clay based, conforming respectively to Parts 1 and 2 of IS 1489 : 1991, involves the addition of 'pozzolana' (flyash or calcined clay) - a mineral additive containing silica; the pozzolana is generally cheaper than the cement it replaces. These cements hydrate and gain strength relatively slowly, and therefore require curing over a comparatively longer period. They are suitable in situations (such as mass concreting) where a low rate of heat of hydration is desired. Hydrophobic Portland Cement (HPC) - conforming to IS 8043 : 1991, is obtained by inter-grinding Portland cement wit11 0.1-0.4 pmcent of oleic acid or stearic acid. The 'hydrophobic' (water-resistant) property is due to the formation of a water-repellent film around each particle of cement. During the mixing of concrete,

' The term 'characleristic strength' is defined in Section 2.6.1. Higher grade OPC is now widely available in India, and is achieved in cement manufacture by increased proportion of lime (which enhances C3S) and increased fineness (up to 325 kg/m2). The higher the grade of cement, the quicker will be the strength gain of the concrete mixture. However, in the long run, the strength developmnt curves mare or less converge for the various grades of cement.

this film is broken, thereby makirlg- it possible for normal hydration to take place. . Although its early strength is low, this cement is suitable in situations where cement b a-~ are s reauircd to be stored for a vroloneed oeriod under unfavourable conditions. because it deteriorates very littlc.

Low Heat Portland Cement (LHPC) - conforming to IS 12600 : 1989, is Portland cement with relatively lowcr contents of thc more rapidly hydrating compounds, C3S and C3A. The process of hydration is slow (as with ~ ~ C ) , . a nthk d conseauent rate of heat generation is also low. This is desirable in mass concretine of gravity dams; as otherwise, the excessive heat of hydration can result in serious ~. cracking. However, because of the slower rate of strength gain, adequate precaution should be taken in their usc such as with regard to rcmoval of formwork, etc.

-

-

Sulphate ~esisting Portland Cement (SRPC) - conforming to IS 12330 : 1988, is Portland cement with a very low C3A content and ground finer than OPC. This cement is 'sulphatc-resistant' bccausc the disintearation of concrete, :auwd hy the, I C ~ U I M , . Hence, the section would have 'cracked'. The cracked-trausforn~edsection is shown in Fig. 4.1 1. Transformed Section Pro~erties: e

.

modular ratio in = 13.33 (for M 20 concrete) Transformed steel area = mA,, = 13.33x1963 = 26167 nun' Equating moments of areas about the neutral (centroidal) axis, 300(kd)' = 26167(550 - kd) 2

4.6.2 Permissible Stresses In the traditional working stress method, analysis requires the designer lo verify that the calculated stresses [Eq. 4.19 and 4.201 under service loads are within 'pe~missible liinirs'. The 'pennissible stress' in concrete under flexural compscssion (dcnoted as Ucb, by the Code) is as given in Table 4.1. The 'permissible stress' in tcnsion steel GS,(specified in Table 22 of the Code) takes values of 140 MPa, 230 MPa and 275 MPa for Fe 250, Fe 415 and Fe 500 grades respecti;ely. However, lor Fe 250 grade, the permissible strcss is reduced to 130 MPa if the bar diamncter cxcecds 20nrm.

BEHAVIOUR IN FLEXURE

H 6 REINFORCED CONCRETE DESIGN

In the case of reinforcing steel under con~pressionin flexural members, the permissible stress u, is limited to the culcdnrcd corupessive stress in the surrourrrling concrctc nnultiplierl by 1.5 times the rt~orlrrlnrmfio or. 6,, (maximum permissible compressive stress in steel given in Table 22 of the Code), whichever is lmvert. The specified values of 5, are 130 MPa, I90 MPa and 190 MPa for Fe 250, Fe 415 and Fe 500 grades respectively.

u,, = m~~~~ x

1-k,

k6

The product n1Ucb, is a constant, equal to 2801 3, according to the Code [Eq. 4.91, whercby the above equalion can be solved to give:

',

k, =

280 280 + 30,,

(4.23)

where os, is in MPa units

4.6.3 Allowable Bending Moment When it is desired to compute the 'allowable bending moment' capacity of a beam of known cross-section, in accordance with WSM, the procedure to be adopted is very similar to that given in Secrion 4.6.1 and Example 4.2. Here, the stresses in concrete and steel (f, andf,,) are taken as their respective 'pcnnissible stresses' (occand Us,)as specified in Section 4.6.2. Considcring the momcnt with reference to the tension steel [Fig. 4.101,

Considering the moment wtth reference to thc compression in concrete [Fig. 4.101,

M,,, = 0.5 o , ,

b(kd)(jd)

(4.22b)

In a given beam section, the permissible stresses in both steel and concrete may not he reached simultaneously. Hence, the lower of thc two mome~fscomputed by Eq. 4.22a and 4.22b will give the correct permissible moment, and the cotresponding stress (eitherf,, orf,) will be the onc to reach thc permissible limit. Alternatively, with the knowledge of certain constants (discussed in thc subsection to follow), it is possible to predict whether it is thc steel or thc concrete that controls

Krr. 'Balanced (WSM)' section constants

(a) beam Section

OALWCEO SECTION

(d)

In the working stress method, the 'balanced' section is one in which both tensile steel stress h.,and maximum compressive stress f, simultaneously reach their allowable limits ox,and ocb, respectively [Fig. 4.121 under service louds. The corresponding area of steel A,, is denoted as A,,,b; the percentage rcinforcenlent p, 100A,,,d bd is denoted as p,,,; the neutral axis depth factor is dcnotcd as kb; the lcver arm depth factor is denoted as j,; and the allowable moment o l the section is denoted as M,,. For such a case, from thc linear distribution of stresses [Fig. 4.12(c)l in tbc transformed-cracked section [Eq. 4.211, it follows that:

-

(b) transformed section

UNDER-REINFORCE0 SECTlON

z,, and

(c)

stresses

0YER.REINFORCED

SECTION

comparison with under- and over-reinforced sections Fig. 4.12 'Balanced (WSM)' section

Further, considering the equilibrium of forces C=T,it follows that

0.5a,,b(k6d) = A,, o,,

Finally, considering moment equilibriun~,

'Generally, the value of 1.5,~ times the cnlculated compressive stress is lower than hence controls

117

BEHAVIOUR IN FLEXURE

120 REINFORCED CONCRETE DESIGN !

withp, for p, < P , , ~ .Fu~ther,although the rate of gain In M,a ikreases with the use of higher strength stcel, the 'balanced' section limit IS leached at a lower perccntage of stcel.

121

Finally, it may be noted that 'over-reinforced (WSM)' sections often may turn out to be ander-reinforced with reference to the ultimate limit stare (leading to ductile failure), except when the percentage of tension steel is very high. For example, when M 20 concrete and Fe 415 steel are used in a beam section, if the tension steel arca exceeds 0.439 percent, by the w o r ~ n gstress method the section i s 'over-reinforced (WSM)', -but it is under-reinforced in the ultimate limit sense (up to tension steel area of 0.961 percent) [refer Section 4.71.

Analysis Aids The variation of ~ , ~ / b dwithp, ' for different grades of concrete and steel (depicted in Fig. 4.13) is expressed in tabular form and presented in Tables A.l(a) and (b) in Appendix A of this book. These Tables serve as useful analysis aids, enabling the rapid determination of Marlfor any given singly reinforced rectangular beam section. The use of these Tables is demonstrated in Example 4.3. EXAMPLE 4.3

Consider the same beam section [Fig. 4.11 of Examples 4.1 and 4.21. Assuming M 20 grade concrete and Fe 415 grade steel, determine the allowable bending moment, and the stresses in concrete and steel corresponding to this moment. SOLUTION

Given: oCbc= 7.0 MPa, a,, = 230MPa, m = 13.33.A,, = 1963 mml, b=300mm,d=550mm

0.00

0.25

0.50 0.75 1.00 percentage tension steel (pt)

1.25

The transformed section properties [Fig. 4.1 1(b)] have already been worked out in Example 4.2. Accordingly, kd = 234.6 mm =$ k = 0.4265. The neutral axis depth factor kb is a constant [Eq. 4.231.

1.50

For Fe 415 steel (a*, = 230 MPa), k

Fig. 4.13 Variation o f , ~ ~ a l with b B pl for different grades of concrete and steel

s&&&s: As k

l:l.',,.,

I.,,'>'

!$': , ' p,,b (i.e., for 'over-reinforced (WSM)' sections), the rate of gain in allowable moment capacity with increase in tensile reinforcement area drops off rapidly. This is so, because the allowable limit of stress is rcached in concrete in compression, and, unless the compression capacity is suilably enhanced', there is not much to gain in boosting thc flexural tensile capacity of the beam section-either by adding more tension steel arca or by made of stcel. . improvinn . - the ..

-

"280 +2803(23O) =0.2887

> kh. the section is 'over-reinforced (WSM)'. [Alternatively,

100x1963 = 1.190. P , , ~= 0.440 for M 20 concrete with Fe 415 steel. As

" = 300x550

p, > P,,~,the section is 'over-reinforced (WSM)'].

.

Accordingly, the concrete

= 7.0 MPa (for M 20 concrete). stress controls, and f , = aebr Applying T = C,

For this 're&on; 'over-reinfo~cad.:[W~~)': beams ' d r ~considare~.to:tj6.:Hl$hiY. unecondmical~inlhe.tfadil;onal WSM method ol des'gn:. ; ; ~,, .:;. . _.,-.a ~:,.

.

.., . ,

1: li, I,wn , '? ,.

D ~the,assumption kd 5 D ~is , incom~ect. Now solving Eq. 4.40 for k d 2 D,, kd = 181.8 mm

k = 181.81 520 = 0.3496

For a 'balanced (WSM)' section with g,, = 130MPa [Eq. 4.231, kb=a4179 As

k =0.3496< kb, the section is 'under-reinforced (WSM)' whereby

f,,, = D,, = 130 MPa. Considering the linear stress distribution [Eq. 4.32(b)], 130

The formula for M,,b is obtainable by Eq. 4.25 For the present problem, applying the various fo~mulae with k b = 0.418, b = 3 0 0 mm, d = 550 mm, m = 13.33,0, = 130MPa, bCbc =7,0MPa, and ~ = 1 7 5 ~ 1 0 ~ ~ m m , M, =0.5x7.0~300x(0.4179x550)~(1-0.4179/3)550

.

f =-x+&.!-

13.33 520-181.8

Substituting in Eq. 4.42.

MOn= 226 kNm

[Eq. 4.251

= 1 1 4 . 2 7 ~ 1 0Nmm. ~ [Eq. 4.451 [Eq. 4.461

fcsc

=7.0x(1-

)

50 = 5.477 MPa 0.4179~550

[Eq. 4.441

Note that allowance has to be made for the area A,, displaced by the concrete; this is done in thecalculation for the required area of the compression steel (Eq.4,471.

'refer Section 4.6.2

= 5.24 MPa

(< oc, = 7.0

MP~)

134 REINFORCED CONCRETE DESIGN

4.7 ANALYSIS AT ULTIMATE LIMIT STATE

BEHAVIOUR IN FLEXURE

135

beam is u n d e r - r e i r f i d or over-reinforced, collapse invariably occurs by the crushiug of concrete (as explained in Sectiou 4.5.3).

Whereas the previous section (Section 4.6) dealt with the 'analysis at sevvice loads', the present section deals with the 'analysis at ultbnate loads'. The former is based 011 the rvwking stress method (WSM) and is also applicable to the analysis of 'serviceability limit states', whercas the latter is based on thc 'ultimate limit state' oC the limit .stat& nrerhod (LSM). Havine studied analvsis at 'service loads' in somc detail ( i n c l u h g the solution to a number of Example problems), it is likely that the student may get somewhat confused while undertaking the task of analysing the same problems at 'ultimate loads'. The important question that is likely to disturb the student is - why go t l ~ o u g hthis process of analysing at service loads as well as ultimate loads? The answer to this question was given in Chapter 3, where it was explained that a structure has to be both safe (at various ultimate limit states) and serviceable (at various setviceability limit states). At ultimate iimit states, the loads are those corresponding to impending failure of structure, whereas at serviceability linlit states, the loads and stresses are those applicable in the day-to-day service of the structure. This section investigates the 'safety' of flexural members (of given design) at thc ultimate limit state in flexure. The previous section discussed the calculatim~of flcxural stresses under service loads required for serviceability analysis (described in Chapter 10). and also the calculation of 'allowable bending moment' based on the WSM concept of permissible stresses. The latter was included to enable the student to gain a first-hand understanding of the traditional (and, earlier much-used) working stress method which retains a place in the Code, albeit as an Appendix, and is sonletimes used in the design of special structures such as water tanks and mad bridges. Therefore, the studcnt will do well to keep in perspective the background of the present section, dealing with the analysis at the 'ultimate limit state in flexure'. The expressions derived here will find use again in the next chapter (Chapter 5). which deals with the design of reinforced concrete beams at the ultinlate limit state in flexure. In this section, the Code procedure for analysis is discussed. The calculations are based on the idealised stress-strain curves for concrete and steel, as specified by the Code. Moreover, the design stress-strain curves (involving partial safety factors y , , y,) are used, as explained in Section 3.6.

In the casc of mild steel, which has a well-defined yield point (E, = 0.87f,/Es, as shown in Fig. 3.6), the requirement (0 cited above may appear to be conservative. However, the Code specifics a uniform criterion [Eq. 4.481 for all grades of steel. The intention here is to ensure that 'yielding' of the tcnsion stcel takes place at the ulti~nateh u t state, so Illat the coltsequent failure is ducrile in nature, providing ample warning of the impending collapse.

4.7.1 Assumptions in Analysis

4.7.2 Limiting Depth of Neutral Axis

The behaviour of reinforced concrete beam sections at ultimate loads has been explained in detail in Section 4.5.3. The basic assumptions involved in the analysis at the ultimate limit state of flexure (CI. 38.1 of the Code) are listed here [see also Fig. 4.171. (Most of these assumptions havc already been explained earlier.) a) Plane sections normal to the beam axis rcmain plane aiter bending, i.e., in an initially straight beam, strain vaies liuearly over the depth of the section.

Based on the assumption given above, an expressiou for the depth of the neutral axis at the ultimate limit state, A,,, call be easily obtaincd from the strain diagram in Fig. 4.17(b). Considering similar triangles,

b) The maximum compressive strain in concrete (at the outermost fibre)&,,, shall be taken as 0.0035 [Fig. 4,17(b)]. This is so, because regardless of whether the

I t is intelesting to note tbnt llle use of fhe deleripz yield stress f& = 0.87 f l , instearl uf the characteristic yield stress,&,results in a slightly icsser (and hence, less conservative!) value of the yield strain 5 [I&r Pigs 3.6, 3.71.

c)The design stress-strain curve oC concrete in flexural compression (recommended by the Code) is as depicted in Fig. 3.5. [The Codc also permits the usc of any other slrape of the stress-strain curve which ,rsrdts in substantial agreement with the results of tests.] The partial safety factor y,= 1.5 is to be considered. d) The tensile strength of the concrete is ignored. e) The design stress-strain cufves for mild steel and cold-worked bars are as depicted in Fig. 3.6 and Fig. 3.7 respectively. The partial safety factor y, = 1.15 is to be considered.

O The strain

E,,

in thc tcnsion reinforccmcnt (at its centroid) at the ultimate limit

state shall not be less

than^,,* [Fig. 4.271, deiined as: c&

= (0.87 f,/ E x ) + 0.002

(4.48)

This is cquivalcut to defining the yield stress fy of stcel as thc stress corresponding to 0.002 strain offset (0.2% proof stress) - regardless of whether the steel has a welldefined yield point or not. The yield strain coxresponding to fy is then given by 0.002+ f y / E , . Introducing rhe partial safety factor y , = 1.15 to allow for thc variability in the steel strength, the designt yield strength, f, = fY/1.l5 = 0.87h. and using this in lieu o f f , , the yield strain E

&,

[refer Fig. 3.71 is given by:

= 0.002 t (O.87d, /Ex)

E:,

136 REINFORCED CONCRETE DESIGN

BEHAVIOUR IN FLEXURE 137

!'

4.7.3 Analysis of Singly Reinforced Rectangular Sections Analysis of a given reinforced concrete section at the ultimate limit state of flcxure implies the determination of the ultimate moment of resistance MllR of the section. This is tasily obtained from the couple resulting from the flexural stresses [Pig. 4.17(c)]: C".Z = z7;;

M,,,=

(4.51)

wherc C,,and T, are the resultant (ultimate) forces in conipression and tension respectively, and z is the lever arm. (a) beam section

(b)

strains

T,M= f

(c) stresses

(4.52)

d s

where

f., = 0.87fy

Fig. 4.17 Behaviour of singly reinforced rectangular section at ultimate limit

state in flexure .:,I I.'

According to thc Code [rcquircment (f) in Section 4.7.11, &, > &,: , implyillg that thcre is a limiting (maximum) value of the neutral axis dcpth x,,,,,, corresponding to E ~ ,= &,: . This is obtained by substituting the expression for Eq. 4.49:

>:..

,,.i

I:!

i i:

LI; L! 1 !;:is

&,:

[Eq. 4.481 in

The values o f ~ , , , , ~ , , for ~ / ddifferent grades of steel, obtained by applying Eq. 4.50, are listed in Table4.3. It may be noted that the constillla given in TabIe4.3 are applicable to all cross-sectional shapes, and remain valid for doubly reinforced sections as well.

and the line of action of steel.

T, corresponds to the level of

far 4, < x,r,,nax the centroid of the tension

Concrete Stress Block In Compression In order to determine the magnitude of C,, and its line of action, it is necessary to analyse the concrete stress block in compression. As ultimate failure of a reinforced concrete beam in flexure occurs by the crushing of concrete, for both under- and overreinforced beams, the shape of the compressive stress distribution ('stress block') at failure will be, in both cases, as shown in Fig. 4.18. [also refer assumptions (b) and (c) in Section4.7.11. The value of C,, can be computed knowing that the compressive stress in concrete is uniform a1 0.447 f& for a depth of 3x,, / 7, and below this it varies parabolically over a depth of 4x,, 17 to zero at the neut~alaxis [Fig. 4.181.

Table 4.3 L~mltlngdepth of neutral axls tor diflerent grades of steel

if. :t., :!j

.

i4

It

,

The limiting depth of neutral axis x,,,,!~~ comsponds to the so-called balanced section, i.e., a section that is expected to result in a 'balanced' Iailure at the ?iltimate limit state in flexurr. [refer Section 4.5.31. If thc.neutra1 axis depth x,, is less than x,,,,,,,, then the section is rcnde~reinforced(resulting in a 'tension' failure); whereas if x,,exceeds x,,,,,,nr, it is ove~reinforced(resulting in a 'compression' failure).

BEAM SECTION

(truncated)

STRAINS

STRESSES Fig. 4.18

Concrete stress-block parameters in compression For a rectaneular section of width b.

138 REINFORCED CONCRETE DESIGN

Also, the line of action of C, is determined by the centroid of the stress block, located at a distance ? . from the concrete fibres subjected to the maximum compressive strain Accordingly, considking moments of compressive forces C,,, CI and C2[Fig. 4.181 about the maximum compressive strain location,

give11 by step (4): athenvise, repeat steps (2) to ( 5 ) with an improved (say, average) value of x,,, until convergence.

Ultimate-Moment of Resistance The ultrntate moment of fesrstance ME,"of a given beam sectlon 1s obta~nablefrom z, for the case of the singly reinforced rectangular section [Fig. 4.17(d), Fig. 4.181 is g~venby

Eq. 4.51. The lever arm

z = d-0416~"

Solving, i =0.416x,

(4.54)

Depth of Neutral Axis For any given section, the depth of the neutral axis should be such that C,,= satisfying equilibrium of forccs. Equating C,,= T,, with cxpressions for C,, and given hy Eq. 4.53 and Eq. 4.52 respectively:

.,

=

For the condition x,, > x,,,,

0.87f,A,, 0.362f, 6

_______,valid only if resulting ,

E

,, < E : ,

x,, $ x,,,,,,,,

.

T,

1;,,

Accordingly, in terms of the concrete compressive strength, M , , = 0.362fCkbx,(d - 0.416x,,)

.,... ..,, .

~~~~

for allx. ,

,

Alternatively, in terms of the steel tensile stress, Mm = f d , t ( d - 0 . 4 1 6 ~ , ~ )

(4.55)

(4.59)

- ..... . ......

for all xu

with f,,= 0.87fy for

~

.,...

(4.60) x,,

< x ,,>,

~lmitln Moment ~ of Resistance [Fig. 4.171, implyiilg that, at the ultimate

limit state, the steel would not have 'yielded' (as per the proof stress definitiont forfy) 'and the steel stress cannot be taken as f,/y, = 0.87f,. Hence Eq. 4.52 and therefore Eq. 4.55 are not applicable. When the steel has not yielded, the true location of the neutral axis is obtained by a trial-and-el~or method, called saain coe~l~atibiliry method, involving the following steps:

The limiting moment of resistance M,,,,(,,,of a given (singly reinforced, rectangular) section, according to the Code (CI. G-].I), corresponds to the condiiion defined by Eq. 4.50. From Eq. 4.59, it follows that: .-

M,,,I~,,= 0.362fdx ,,mu ( d - 0 . 4 1 6 ~,,)

(4.61)

1) Assume a suitable initial (trial) value of x,, 2) Determine E,, by considering strain compatibility [Eq. 4.491: &,-=

0.0035 ( d l x , , - 1)

(4.56)

3) Determine the design rtress htcon'esponding to E , using the design stress-strain curve [Fig. 3.7, Table 3.21. 4) Derive the value of x, corresponding tof,, by considering T , = f,,A,, and applying the force equilibrium condition C,, = T,,,whereby

5)

' In the

Comparc this value of x,, with the value used in step (1). If the difference between the two values is acceptably small, accept the value

case ollow grade mild

steel (Fe 250). which has a sharply defined yield point, the steel would have yielded and reached f, even at a strain slightly Lower than E:, . In such cases, one may find that/,, = 0.87f,even far values ofx,,slightly in excess of x,,,,,,.

The values of the non-dimensional parameter Kfor different grades of steel [refer Table 4.31 are obtained as 0.1498. 0.1389 and 0.1338 for Fe 250, Fe 415 and Pe 500 respectively.

Limiting Percentage Tensile Steel Corresponding to the li~niting moment of resistance M,,,,~,,,,there is a limiting percentage tensile steel p ,,,, = 100xA,,,,,, p d . An expression for p ,,,,,, is obtainable from Eq. 4.55 with x , = x ,,,, :

,

140 REINFORCED CONCRETE DESIGN

The values of p ,,,,,,, and M ,,,,,,/ b d 2 (in MPn units) for, different combinations of steel and concrete grades are listed in Table 4.4. These values correspond to the socalled 'balanced' section [refer Section 4.5.31 for a singly ~einforcedrectangular section ' singly reinforced rectangular Table 4.4 Limiting values of P,,rw, and ~ , , , , ~ , , , / b dfor beam sections for various grades of steel and concrete.

(b)

M,,,,~J~B values (MPa)

percentage p, shonld not exceed P,,,~,,,a n d the ultimate moment of resistance M,(R should not exceed M,,,li,,, [Table 4.41. The Code (CI. G-l.ld) clearly states:

The topic of design is covered in detail in Chapter 5. The present chapter deals with analysis - and, in analysis, it is not unlikely to encounter beam sections (already constructed) in which p, > p,,,, , whereby x,, > ,,, and M,,# > M ,,, . Evidently, in such 'over-reinforced' sections, the strength requirement may be satisfied, but not the ductilityt requirement. The question arises: are such sections. acceptable ? The answer, in general, would be in the negative, except in certain special situations where the section itself is not 'critical' in terms of ductility, and will not lead to a brittle failure of the strncture under the given factored loads. In such exceptional cases, where M,,R > M,,,and inelastic flexnral response* is never expected to ozcur under the given factored loads, over-reinforced sections cannot be strictly objected to. It may be noted that the exact determination of M,,Rof an over-reinforced section generally involves considerable computational effort, as explained in the next section. An approximate (but conservative) estimate of the ultimate moment capacity of such a section is given by the limiting moment of resistance, M,,,,j,,,,which can be easily computed.

Variation of M

For x,,< x,,,,,

wlth pt (for singly reinforced rectangular sections)

. it is possible to arrive at a simple closed-form

expression lor the

ultimate moment of resistance of a given section with a specified p, 5 p,,li,. First, expressing A,, in terms of p, :

Safety at Ultimate Limif State in Flexure The bending moment expected at a beam section at the dtirnute limit state due to the fizctored loads is called the factored momer,t M,,. For the consideration of vzious combinntions of loads (dead loads, live loads, wind loads, etc.), appropriate load factors should be applied to the specified 'characteristic' loads (as explained in Chapter 31, and the factored moment M,, is determined by structural analysis. The beam scction will bc considered to be 'safe', according to the Code, if its ultimate moment of resistnnce M,,Ris greater than or cqual to the factored moment M,,. In other words, for such a design, theprobobilily of fnilure is acceptably low. It is also the intention of the Code to ensurc that nt ultimate failure in flexure, the type of failure should bc a tewsion (ductile) failure - as explained earlier. For this reason, the Code requires the designer to ensure that x,, 5 x,,,",, [Table 4.31, whereby it follows that, for n s i q l y reinforced rectnng~rlnrsection, the tensilc reinforcement

and then substituting ih Eq. 4.55.

Further suhstitnting Eq. 4.63 and Eq. 4.64 in Eq. 4.60,

'

The ductility requirelllenr may be pmly satisfied in the case of mild steel (Fe 250). evcn if x,, slightly exceeds x,,,,,,,,; this is explained later with referrnee to Fig. 4.19. [See also footnote on

136.1

Pfor details on 'plastic hinge' formation at the ultimate lmit state, refer Chapter 9.

142 REINFORCED CONCRETE DESIGN

BEHAVIOUR IN FLEXURE

143

t, *,, > x , For p, > p,,,,,,,$, whereby the design stress in the tension steel takes a value&, which in general is not a constant, and depends on the value of xr, [Eq. 4.561.

To determine&,, the (trial-and-error) Strain conipatibility metlmd (described earlier) has to be employed. The final expression, comparable to Eq. 4.65, takes the following fonn:

where f,, 2 0.87f, has to satisfy the force equilibrium condition [Eq. 4.571, and the strain t?, corresponding to fs,[Fig. 3.6, 3.71 must satisfy the strain compatibility condition [Eq. 4.561. For convenience, Eq. 4.57 is re-arranged as follows: percentage tension reinforcement (pt ) 400

I

1

The steps involved in the 'stmin-compatibility method for determining ~ , , ~ / b d ' for a given p, , are as follows: I) Assume an initial (trial) value of x,,/d : say, x,,,,,,,, /d ;

2) Determine E , using Eq. 4.56: 3) Detenninef,, from E , using the design stress-strain curves [Fig. 3.6, 3.71; 4) Calculate the new valne of x,,/d using Eq. 4.67; 5) Compare the new value of x,/d with the old value. If the difference is within acceptable tolerance, proceed to step (6); othelwise repeat steps (2) to (5) until convergence is attained. 6) Apply Eq. 4.66 and determine ~ . , / b d ' .

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

percentage tension reinforcement (pd Fig. 4.19 Variation of (a) M d b d ' a n d (b)

f,t

with Pt

4.0

A quick solution can be obtained by means of a computer program. The relationship between M,,,/D~' (expressed in MPa units) and p, is plotted it1 Pig. 4.19(a) for two typical grades of steel (Fe 250 and Fe 415) combined with the commonly used grades of.concrk: (M20 and M 25). The corresponding relationship between stress f , (at the ultimate limit state) andp, is depicted in Fig. 4.19(b). The relatively thick lines represent the under-reinforced condition p, 2 p,,ri,,, (i.e., 'This condition p, > p,,li,, is not permitted in design. Its only relevance is in mralysis.

j

EXAMPLE 4.9

M,,RS M,,R,II,J.whereas the thin lines denote the over-reinforced condition p, > p,,ll,>, (i.e., M,,R > M,,R,IC.),and the transition points are marked by thin vertical lines. It can be seen that these curves in Fig. 4.19(a) (for the ultimate limit state) bear resen~blancewith the corresponding curves in Fig. 4.13 (for the service load state).

It may be noted from Fig. 4.19(b) that with the steel percentage limited top,~i,,,,as ultimate moment of resistance M,,, is reached, the steel would have already 'yielded' (6, = 0.87f) and gone into the domain of large inelastic strains, thus ensuring a ductile response. For p, > pt,~i,,~, the tension steel would not have 'yielded' at the ultimate limit state, with the definition of stecl strain at balanced condition as E ; = 0.002 + (0.87f,)/Ex [refer Section 4.7.11. Howcvcr, in the case of Fe 250 steel [Fig. 3.61, it can be seen that 'yielding' will actually take place even with a steel strain less than E:,

~~~~~~~i~~ the neutral axis depth x , (at the ultimate limit state) for the beam section in Example 4.2.

.. .

SOLUTION

G~v~,,:b = 300 -, d = 550 -,A,,= 1963 mm2,fy= 415 MPa .fci=20 MPa For Fe 415 stee1.x , 0 . 2 for D , / d 5 0.2

glven by

C,,,, = 0362fd,,*,,

condttion of Df /d 5 0.2 for all grades of steel - to replesent the cond~tion

0 . 1 5,,,,,, ~ +0.65Df

IS

- 250 )x 100

= 276 1 nun,

1;' = 0.87 x 250x 3695 = 803662 N. Applymg the f o ~ c equilibrium e condmon 1810x,,+ 536400= 803662

(4.76)

=$ x ,

(c,~,,+ C,,,

= I;,),

= 147.7 n u n < X u f =233 3 m m .

Hence, tlus calculated valuc of x,, is also not correct. As D~ < xc,,,) is that the estimation of y/ is made somewhat simpler. Of course, for .q,,, 5 DJ (i.e., neutral axis within the flange),

As mentioned earlier, when it is found by analysis of a given T-section that , then the strain compatibility method has to be applied. As an may be taken as M,,,li,?,, given by approximate and conservative estimate, MGrR Eq. 4.76 1 4.77. From the point of view of design (to be discussed in~chapter5), M,,,J~,,,, provides a measure of the ultimate moment capacity that can be expected from a T-section of given proportions. If the section has to be designed for a factored moment Mu > M,,,,o,,, then this calls for the provision of comp~essionreinforcement in addition to extra tension reinforcement.

= 536400 N.

the depth yf ( 5 D f ) of the equwalent conclete stress block

IS obtained

y,,

x,, >x,,,,,,,

.

=0

as: 15x. +0.65Df =(0.15~,,+65) nun.

a 1810x,, +(804.6x,, + 348660) = 803662. a x,, = 174.0mm < x ,,,,,,,, ; hcnce, the assumption f,,= 0.87f, is OK. J y f = (0.15 x 174.0) + 65.0 = 91.1 mm Taking moments of C,, and C,, about the centroid of tension steel,

EXAMPLE 4.12 Determine the ultimate moment of resistance for the T-section in Example 4.4

.

EXAMPLE 4.13

SOLUTION

.

Given: bf= 850 mm, DJ = 100 mm, b,, = 250 nun, d = 520 mm, A,, = 3695 mm2, f,= 250 MPa and fck = 20 MPa x,,,, / d = 0 , 5 3 1 f o r F e 2 5 0 = $ x , =0531x520=276.1mm First assuming x,, 5 D, and x,, 5 x,,, , and considering force equilibrium

C. = T,,

0.362fckb,x,, = 0.87fYA,,

= = 130.6 mm > D,, = 100 mm. 0.362~20~850 Hence, this calculated value of x,, is not correct, as X,, > Df =)

0.87x250x3695 x,, = -

Repeat the T-section problem 6 - 280 bars SOLUTION

.

111 Example

4.12, consldenng 8 - 280 bars Instead of

Glven: by= 850 mm, DJ= 100 rmn, b,, = 250 mm, d = 520 mm,f, = 250 MPa and

hk = 20 MPa, A , = 8 5, (28 = 4926 mm2 4 x,, = U 6 . 1 nun (as in Example 4.12) 2

BEHAVIOUR IN FLEXURE 153

152 REINFORCED CONCRETE DESIGN

.

First assuming x,, 2 Dl and x,, 5 x

, ,.

'F

Hence this calculated value of x,, is not correct

Approxinlate Solution An approximate and conservative solution for M,n can be obtained by limiting x,, to x ,,,, = 276.1 mm,and taking moments of C,,, and Crf about the centroid of the tension steel (Note that, following the Code procedure, D,/d = 100/520 = 0.192 < 0.2 J y , = Df = 100 mm [Eq. 4.761 ). . Accordingly,

As x,, > Dl,

.

.

C,,,, = 0.362 x 2 0 x 250x,, =(1810x,,)N. Assuming x,, 2% D, = 233.3 mm,

c , =~0.447 x 20 x (850- 250) x 100

. x, .

= 536400 N. Further assuming x,, -< x,,,,n,, = 276.1 m,

4.7.5 Analysis of Doubly Reinforced Sections

= 0 . 8 7 ~ 2 5 0 ~ 4 9 2=6107140.5 N.

-

Applying the force equilibrium condition x,, =

(c,,,,, + C,(

=q r ) ,

1071405 - 536400 = 295,6 1810 which implies x,, > x D f = 233.3 mm, but not x,, 2 x,,,, = 276.1 mm

Doubly reinforced beam sections (i.e., sections with compression steel as well as tension steel) were introduced in Section 4.6.5, where the analysis at service loads was discussed. The present section deals with the analysis of these beam sections (rectangular) at the ulfirnatelimit stale.

Exact Solution (considering strain conrpofibility) Corresponding to x,, = 295.6 m,

E,, = 0.0035(520/295.6-

1) = 0.00266

[Eq. 4.561 which is clearly greater than the strain at yield far Fe 250, i.e., 0 . 8 7 ~ 2 5 0 / ( 2 . 0 ~ 1=0 ~0.00109. ) Hcnce, the design stccl stress is indeed f,, = 0.87fy , and the so calculated x,, = 295.6 nun is the correct depth of the neutral axis' . Accordingly. M,,,= C ,,,, ( d - 0 . 4 1 6 ~ , , ) + C , ~ ( d - D ~ / 2 ) = ( 1 8 1 0 ~2 9 5 . 6 ) ~(52O-O.4l6~295.6)+53640Ox(S2O-50) = 4 6 4 . 5 ~ 1 0Nmm ~ =465kNm > A4,,,,,, This is the collect estimate of the ultimate moment capacity of the section; as the steel strain is beyond the yield strain a Limited amount of ductile behaviour can also he expected. However, as per the Code, this will not qualify as an admissible underreinforced section since x,, > x,,,. [Note that if the &, computed had turned out to be less than E ~ f,, , < 0.87fy and a trial-and-error procedure has to be resorted to.]

(a) beam section

(b) strains

a case where, heing Fe 250 grade steel with a sbnrp yield point, the strain at first yield. &, = f, / E , , is lower than the suain for the 'balanced' condition E:, specified by the Code. , .,

(d) resultant force

Fig. 4.21 Behaviour of doubly reinforced rectangular section at ultimate limit state The distributions of stresses and strains in a 'doubly reinforced' rectangular section [Fig. 4.211 are similar to those obtained in a 'singly reinforced' section [Fig. 4.171, except that there is a stressf, in the compression steel (area A,,) which also needs to be accounted for. This stressf,, may or may not reach the design yield stress 0.87f,., depending on the strain E,, in the compression steel. An expression for E , can be easily obtained from strain compatibility [Rg. 4.21(b)I: E,

'This

(c) stresses

dy1,)

= 0.0035 ~ ( -l

(4.78)

19

Hence, even though x. > Fig. 4.19(b).

I,,,,,,,,the

steel has yielded. See also footnote on p. 136 and

where d' is the distance between the centroid of the compression steel and thc extreme compression fibre in the concrete. In practice, the ratio d'/d is f0ur.d to

154 REINFORCED CONCRETE DESIGN

BEHAVIOUR IN FLEXURE

vary in the range 0.05 to 0.20. It can be shown that the compres.sion'stee1 will, in most cases, attain the design yield stress (f,, = 0.87fy) in'the case of Fe 250 grade steel, but is generally unlikely to do s o in the case of Fe 415 and Fe 500 (because of their higher strains at yield). Values of the stress& (comsponding to x,, = x,,,,) for various grades of steel and ratios of d'/d are listed in Table 4.5.

155

Limiting Moment of Resistance The 'limiting' value of M,,R, obtained for the condition x,, = x,,,,,~,, is given by the following expression :

= x,,,,,~~, - for various d'/d ratios and different grades of compression steel

Table 4.5 Value o f f , (in MPa units) at x.

where the value o f f , depends on E , (obtainable from Eq. 4.78 and Table 3.2). For convenience, the values of lhe stress f,, (corresponding to X I , = x , , , , d for various grades of steel and ratios of d'ld are listed in Table 4.5. Linear interpolation may b e used to deternlinef, for any value of d'ld other than the tabulated constants. EXAMPLE 4.14

Determine the ultimatc moment of resistance of the doubly reinforced beam section of Example 4.6. Applymg the condition of force equllibnum [Fig. 4.21(d)]

where C,,cand CttSdenote, respectively, the resultant compressive forces in the concrete and the compression steel. For convenience, the full area of the concrete under compression ( bx xu) is assumed to be effective in estimating C,,,. The force in concrete area displaced by steel (equal to A,, stressed to a level that is exactly or nearly equal to 0.447 fck,the strcss in concrete) already included in C,,,, is accounted for in the estimation of C , as follows:

T , =f,& where&, = 0.87fy if x,, 5 x,,,,,,. xu 1s obtainable from Eq. 4.79 as:

Accordingly, the depth of the neutral axis

SOLUTION

Given : b = 300 nun, d = 550 nun, A,, = 3054 mm2, f,.= 250 MPa and f,k = 20 MPa, d = 50 nun, A,, = 982 mm2 x,,,,,,,/d=0.531 forFc250 3 x,,,,,I, = 0.531~550=292.1 nun. Assuming f,, = f,, = 0.87 f,., and considering forceequilibrium :

C , , + C , , = T,, with C,, = 0 . 3 6 2 ~ 2 0 ~ 3 0 0 x x=,(,2 1 7 2 x , , ) ~ C,,, = (0.87 x 250 - 0.447 x 20)x 982 =204806 N T, = 0.87 x 250 x 3054 = 664 245N =YZl72 x,, + 204806 = 664 245 =Y x,,=211,5nun

3 f _ = 0.87fy is also justified.

E, =

0'87 250 = 0.00109 2x10~

156

BEHAVIOUR IN FLEXURE

REINFORCED CONCRETE DESIGN

EXAMPLE 4.15

Third Cycle :

Repeat the problem in Example 4.14, cons~deringFe 415 instead of Fe 250.

1) x,, = L(323 + 328) = 325.5 mm

. .

2

Given : b = 300 mm, d = 550 mm, A, = 3054 mm2, fy = 415 MPa and j & = 20 MPa, d ' = 50 mm, A, = 982 mm2 ,, /d = 0.479 for Fe 415 =3 x,,,,= 0.479 x550 = 263.5 lm

*

/

MZtR= (0.362x20~300~326)~(550-0.~6x326) +

(0.87 x 415 x 3054) - (0 87 x 415 - 0.447 X 20) X 982 0.362 x 20 x 300 = 348 5 mm > X ,,,",*, = 263.5 mm.

(353.5-0.447xZO)x982x

x. =

f,

Firs1 Cycle : 263.5 mm < x,, < 348.5 mm. Assume xm = !. (263.5 + 348.5) = 306 mm.

1) Evidently,

2)

2

3) [Eq. 4.781 3 E,, = 0.0035(1-501306) = 0.00293 Es, = 0.0035(550/306- 1) = 0.00279 4) [Eq. 4.561 [Table 3.21 =$ f,, = 35 1.8+ (360.9 - 351.8) x (293 - 276)/(380 - 276) 5) = 353.3 MPa and f,, = 351.8 + (360.9 - 351.8) x (279 - 276)/(380-276) = 352.1 MPa 6) =, x,, =(3054x 352.1-982x 353.3+8779)/2172 = 339.3 mm.

*

Second Cycle : 1) Assume x,,

-

1 -(306 2

+ 339)

= 323

*

&, = 0.00296 2) [Eq. 4.781 3) [Eq. 4.561 =) E , = 0.00246 f, =353.5 MPa (converged, insensitive to changes in x,,) 4) [Table 3.21

*

5)

3

and f,= 344.1 MPa x,, =(3054x 344.1-982~353.5+8779)/2172 = 328.0 mm.

*

Taking x,, = 326 mm, and applying Eq. 4.82,

Assuming f,,=Ar = 0.87 xfy, and considering force equilibrium [Eq. 4.811,

Exact Solution (considering strain compatibility) : x3054-(f, -0.447~20)x 982 Applying [Eq. 4.811 : x,, = 0.362 x 20 x 300

*

2) [Eq. 4.561 E., = 0.00241 f,, = 342.8 MPa 3) [Table 3.21 4) x,, =(3054x 342.8- 982x353.5+8779)/2172 = 326.2 mm (converged)

SOLUTION

Evidently, the section is over-reinforced

157

(550-50)

= 4 6 2 . 6 ~ 1 0~~ m =m463 W n (Note : this moment is associated w ~ t hbrrnle failure).

.

Approximate Solution AS an E,

approximate and conservative estimate, limiting XU to XU,,~G = 263.5

= 0.0035(1- 501263 .5) = 0.00284

=?f, = 352.5 MPa [Table 3.21. [This value is alternatively obtainable from Table 4.5 for d ' l d = 0.09 and Fe 415.1 Accordingly, limiting the ultimate moment of resistance M,,R to the 'limiting moment' M,.,i,>,[Eq. 4.831,

M,,J~,,~ = 0.362 x 20 x 300 x 263.5 x (550-0.416 x 263.5) + (352.5-0.447 x 20) x 982 r (550-50) = 420.7 x106 Nmm = 421 kh'm. EXAMPLE 4.16

Determine the ultimate moment of resistance of the doubly reinforced section shown in Fig. 4.22. Assume M 20 concrete and Fe 415 steel.

158

.

BEHAWOUR IN FLEXURE

REINFORCED CONCRETE DESIGN

4.7.6 Balanced Doubly Reinforced Sections

SOLUTiON Glven : b = 300 mm, d = 655 mm, d' = 45 nun,& = 415 MPa and fck = 20 MPa Arc =

x,,

. .

n ( i 5 ) x x 2 = 491

2 =982 m2, A,, = 491 x 4 = 1964 m d

/ d = 0.479 for Fe 415

x,,.

= 0.479 x 655 = 313.7 mm

Assurning (for a first approximation)f,, =& = 0.87&, C,,,= 0.362 x 20 x 300 x x,, = ( 2 1 7 b ) N C , ,= (0.87 x 415 - 0.447 x 20) x 982 = 345772N T,, = 0.87 x 415 x 1964 = 709102N Considering force equilibrium : C,, P:; hence,

the beam is 'under-reinforced'.

However, in Example 4.15, p:works out to 0.914

whilep, provided remains at 0.595; hence,p, < p: L d the beam is 'over-reinforced'. In the case of a 'balenced' section an expression for M, d,>,,,+ effective cover, and should be expressed in rounded figures (for ease in formwork consu~uction). Multiples of 50 mm (or 25 mm) are generally adopted in practice. However, as explained earlier, the

'

effective cover (in nun) may be taken as 40, 50, 65, 70 and 95 respectively for mild, moderate, sever%very severe and extreme conditions of exposure. This will include the contribution of the self-weightof the flexural member. A conservative estimate of the size of the member may be made at the initial stage, for calculating self-weight. The unit weight of concrete should be taken as 25 ldrllm3 [CI. 19.2.1 of Code; see also Appendix B.I of this book].

where, as mentioned earlier, R = M,, /bd2 . It is possible to calculate (A&d directly, without having to determine xJd. By rearranging Eq 5.6,

which is a quadratic equation, whose solution gives:

184

REINFORCED CONCRETE bESlGN

DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE

185

The actual spacing prov~dedshould be rounded off to the nearest lower multiple of 5 mmor 10mm. For conveniencc, Tables A.5 and A.6 (provided in Appendix A) may be referred to - for a quick selection of bar diameter and numberlspacing of bars. The values of bar amns given in Table 5.4 are also obtainable from Table A.5. Table 5.4 also gives the mass per metre length of the bars which may be useful in cost estimation.

The above formula provides a convcnient and dircct estimatc of the area of tension reinforccment in singly reinforced rectangular sections.

Alternative: Use of Design Alds In practice, this is the most widely used method. Expressing the relationship between R = ~ , , / b d 'andp, [Eq. 5.121 in the form of charts or tables for various combinations offy and fck is relatively simple. These are available in design handbooks such as SP : 16 [Ref. 5.51. The tabular format is generally more convenient to deal with than the Chart. Accordingly, Tables A.3(a) and A.3(b) h a w bccn developed (based on Fq.5.12) for M 20, M 25, M 30 and M 35 grades of concrete, each Table covering the three grades of steel [Fe 250, Fe 415 and Fe 5001; these Tables are placed in Appendix A of this book. For a given value of R, and specified values offy and&k, the desired value of p, can be read olf'(using linear interpolation for intermediate values).

5.5.3 Design Check for Strength and Deflection Control The actual A,, and d provided should be worked out, and it should be ensured that the consequent p, is less than p,,,i,>z(for ductile failure at the ultimate limit state). It is good practice to calculate the actual MjIRof the section designed (using Eq. 4.65 or 4.66), and thereby ensure that the actual MuR 2 MI? A check on the adequacy of the depth provided for deflection control is also called for in flexural members. In the case of bcams, the limiting (114ratio given by Eq. 5.5 is generally more-than-adequately satisfied by singly reinforced sections. However, in the case of slabs, the criteria for deflection control are generally critical. In anticipation of this, it is necessary to adopt a suitable value of d at the initial stage of the design itself, as explained in Section 5.5.1. The section should be suitably redesigned if it is found to be inadequate.

Converting Area of Steel to Bars The calculated area of steel (A&,, has to be expressed in terms of bars of specified nominal diameter @ and number (or spacing). Familiarity with the standard bar areas (Ab = ~ @ ~ /[Table 4 ) 5.41 renders this task easy.

EXAMPLE 5.1

Table 5.4 Standard bar areas (Ab = ~ @ ~ and / 4 )mass per metre (kglm)

,

A rectangular reinforced concrete beam, locatcd inside a building in a coastal town, is simply supported on two masonry walls 230 mm thick and 6m apart (centre-tocentre). The beam has to carry, in addition to its own weight, a distributed live load of 10 MVm and a dead load of 5 N / m . Design the beam section for nlaximum moment at midspan. Assume Fe 415 steel.

SOLUTION

For a chosen bar diameter 0, the nu'mbcr of bars rcquired to provide the area of tension steel A,, is given by A,&, taken as a whole number. Alternatively, for a chosen number of bars, the appropriate bar diameter can be worked out. In gome cases, it may be cconornical to sclect a combination of two differcnt bar diameters (close to each other) in order to arrive at an area of steel as close as possible to theA,, calculated. As explained earlier, in the case of slab, the area of steel is expressed in terms of centre-to-centre spacing of bars, given by

The beam is located inside the building, although in a coastal area, and thereby protected against weather, and not directly exposed to 'coastal environment". Hence, according to the Code (Table 3). the exposure condition may be taken as 'moderate'. The corresponding grade of concrete may be taken as M 25 and the clear cover as 30 mm. This cover will be adequate for normal fire resistance requirement also. Deter~niningM,, for design Assume a trial cross-section b = 250 mm, and D = 600 mm (spanllo). Letd=D-50=550mm.

.

'Had the beam been located in the roof, the exposure condition would be 'severe'. Further, if thc stmcture is located at the seafront (subject lo sea water spray), the exposure condition would be 'very severe', according to the Code.

DESIGN

186 REINFORCED CONCRETE DESIGN

;.Effective span (CI. 22.2 of Code) (distance between supportsl =i6.0 m (6.0.-0.23j +0.55 = 6,32(clearspan + d ) Taking the lesser value (as per Code), 1 = 6.0m

,

. . ;. .

Distributed load due to self-weight A W , = 25kN/m3x0.25 mx0.6 m = 3.75 kNlm w, =5.0+3.75 =8.75kN/m, w, = 10.0 kNlm (given) .'.Factored load (as per Code): w,, =1.5(w,,+wLL) = 1.5(8.75+10.0)=28.1kNIm .j Factored Moment (maximum at midspan) M,,~ w , , ~ Z ~ 8 = 2 8 . 1 x 6 . 0 2=1126 8 kNm. Fixing up b, d and D

.

. .

For Fe 415 steel. M,ri,>, = 0.1389 fckbd2[Eq 5.81 For M 25 concrete. M,CJh = 0.1389~25= 3.472 MPa Rib, 7 f,,,= 25 MPa bd Assuming b = 250 mm, for a singly reinforced section, the minimum value of d, corresponding to x,, = x,,>, is given by

BEAMS

AND ONE-WAY

Detailirg Using 3 bars in one layer, 3 x (nd 1 4 ) =I062

\

=$

SLABS FOR

FLEXURE

187

@,*id= 21.2 nnn

Provide 1 -25 @ bru. and 2 -20 @ bars, for which A,, = 491+2(314) = 1119 > 1062. The placement of bars [Fig. 5.21 complies with the clearances specified by the Code.

Desigrr Checks (a) For. sfr.engrh inf7exure Actual d = 450 - 30 - 8 - 2512 = 399 nun. p, =-100x1119 = 1.121 < p , , ~ , ~1.201 ,= 250x399

*

,-

=131.1 x 1 0 b ~ l u>

-

Adopt D = 450 mmr. Assuming 25$ bat's, 8$ stirrups and clear cover of 30 mm, (note that specified cover is required for the stirmps as well), d = 4 5 0 - 3 0 - 8 - 2512=399mrn Deterrrrirtirrg (As&@

OF

M,, = 126 kNm

-Ilcnce,

safe.

4 m:

details of b s curtailmnt are given in Fig. 5.16 details of stinuo reinfffcensnl are gben in Exarrple. 6.1.

Fig, 5.2 Singly reinforced beam design - Example 5.1 [Note: As the actual depth provided (399 mm) is greater than the calculation value ( d = 381 mm), and as the A,, provided (1119 nun') is also greater than the required value (A,,)+ = (1.065 x10-~)x250x399 = 1062 mm2, it. .1s evident

\

. .

,-

.

Solving this quadratic equation in terms ofp, [Eq. 5.121, 100

bd

= 2(415) . % [ , - / ~ ] ; 1 , 0 6 5 ~ 1 0 ~ ~

M

= 3.166 MPa, M 25 bd concrete and Fe 415 steel, p, = 1.065 -which gives the sane result]

[Alternatively, using 'design aids' [Table A.3(a)], for

(without the need for further proof) that the section is safe in flexure.] (h) For deflection corttrol: For p, = 1.121, and

k, = 1.014 (from Fig. 4 of Code or Table 5.2), and, asp, = 0 (singly reinforced beam), kc= 1 =) [Eq 5.51: (Ild) ,,, = 2 0 ~ 1 . 0 1 4 ~=20.28 1 (lld)

'The resulting D h ratio is 1.8, which is satisfactory

=600%99 = 15.04 < (lld),,,

- Hence, OK.

188

DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE

REINFORCED CONCRETE DESIGN

189

EXAMPLE 5.2 Assuming84 bars (A, = a ~ 8 ~ / 4 = 5 0 . 3 m m ' ) , 1000x50 spacing =240 -

Design a one-way slab, with a clear span of 4.0 m, simply supported on 230 mm thick masonry walls, and subjected to a live load of 4 kN1m2 and a surface finish of 2 I kNlm . Assume Fe 415 steel. Assume that the slab is subjected to moderate exposye conditions.

= 208 mm. Max~mumspacing limit. 5 d = 5 X 160 = 800 mm or 450 mm (whichever less) .'.Provide 8 4 @ 200 mm clc for distribution bars.

SOLUTION Delerrni~ri~tg M,, Assume an effective depth d?!!,

Strerrgth check e Providine a clear cover of 30 mm, d = 200-30-1012 = 165 mm

= 160 mm 7. 5.

and an overall depth D = 160 + 40 = 200 m m 4W0 +230 =4230 mm (c / c distance) 4WO+160=4160 mm Taking the lesser values (as per Code), 1 = 4.16 m. Distributed load due to self-weight, Aa,, = 25 w/n13x 0.2 m = 5.0 m/m2 .'.Effective span

:.

= 5 . 0 + 1 . 0 = 6 . 0 k ~ / m ~; w , , = 4 . 0 k ~ / m ~(given) .'.Factored load (as pcr Code) : w, = 1.5(wm + wLL) = 1.5 (6.0+4.0) =15.0 i r ~ / m ' *Factored Moment (maximum at midspan) M,, =~v,,12/8=15.0~4.162/8=32.4 kNmdm. Determining A,, (rnairt bars)

.

.

W

Deflection control check r

M

bd 2

fCk

= 25

3 (As,)reQd= ( 0 . 3 7 4 ~IO-~)XlOOOX 160 = 599 mm21m. [Alternatively, using

'design aids' [Table A.3(a)l, the same result is obtained]. Spacing of bars s = IOOOA, /Ax, Assuming 10@ bars(n, =1rx10'/4=78.5mm'),

=-

Forp, =O.380 and f, =OS8x4I5x-=610 234 ~ l t n m ~ , 628 k,= 1.40 [Fig. 3 of Code or Table 5.21 =$ ( I l d ) ,,, =20X1.40=28.0 ( f l d ),o, =4160/165 =25.2 x,,,,,,,). In order to avoid such a situation, (which is undesirable, and also not permitted by the Code), a correspondingly higher value of A, should by providcd [Eq. 5.17al such that the resulting p, (given by Eq. 4.84).

200 REINFORCED

CONCRETE

DESIGN OF BEAMS

DESIGN

AND ONE-WAY

SLABS FOR FLEXURE 201

I,

Check for Deflection Control satisfied by doubly The limiting (I/d) ratio for deflection control [Eq. 5.51 is reinforced beams, on account of the modification factor (kc)for the compression steel [Table 5.31. However, in the case of relatively shallow beams, a check for deflection control becomes necessary. EXAMPLE 5.4

Design the flexural reinforcement for the beam in Example 5.1, given that its size is limited to 250 mm x 400 mm, and that it has to carry, in addition to the loads already mentioned, a concentrated dead load of 30 kN placed at the midspan point. Assume that the beam is subjected to moderate exposure conditions.

=

Using 3 bars,

.

J T

-- = 27.5 mm

Provide 3 nos 28 nun@ [A,,= 3 x 616 = 1848 mm2]. Actual d(assuming 30 mm clear cover and 8 mm stlrrnps): d = 400-(30+8+28/2) = 348 mm < 350 mm assumed earlier Revising the above calculations with d = 348 mm, = 105 kNm,A,,,ll,= 1.201x250~348/100= 1045 mm2, = 1045+748 = 1793 mm2, (Us,), M ,,,, the section has to be doubly reinforced, with p, > p,,~~,

[tpyx]

parlrrl = 41.61 -

. .

---

= 1.201 for Fe 415 with M 25.

Deterrriirrirrg A,, Considering a 'balanced section' ( x , = x,,,,,,,)

Using 3 b a s ,

@,eqd

=

869/3 L -3

= 19.2 mm

Provide 3 nos 20 mm@[A, = 3 x 314 = 942 nun2 > 869 mm2]. The proposed section is shown in Fig. 5.8. [As an exercise in analysis, the student may verify that this section satisfies the design .I conditions: M,,R 2 Mu and X. 5 Alternative method: using design aids Assuming d = 350 nun, d ' = 50 mm,

A,, =A,,,,;,,, + 4,

1.201 2 where A~,,,,,= -x250x350 = 1051 mn 100 Assuming 20 llvn $ bars for compression steel, d' 48 mm (30 mm clear cover + 8 rnm stirrup + @ 12)

-

Referrine to Table A.4b fM 25 concrete and Fe 415 steel). ,. for d'/d = 501350 = 0.143 and ?4,,/bd2 = 6.073 MPa, by linear interpolation,

P, -2.042

(A,),*,,

2

= ? ! ! ! ? ~ 2 5 0 ~=3 1787 5 ~ mm 100

DESIGN

202 REINFORCED CONCRETE DESIGN

OF

BEAMS

AND ONE-WAY

SLABS

FOR FLEXURE

203

Checkfor deflectiort control p,= 2.124 and is,= 0.58~415~178711848 = 233 MPa e.kt = 0.842 [Table 5.2 or Fig. 4 of Code] kc = 1.263 [Table 5.3 or Fig. 5 of Codc] p, = 1.083 Applying Eq. 5.5, (114 ,>, = 20 x 0.842 x 1.263 = 21.27 (lld) i,,ed= 60001348 = 17.24 < 21.27 -Hence OK.

Provide 3-28 @ for tension steel [A,, = 3x616 = 1848 nun2 > 17871 and 4-16@ for compression steel [A,, = 4x201 = 804 n d > 7991.

,

5.8 DESIGN OF FLANGED BEAM SECTIONS T-beams and L-beams were iritmduced in Section 4.6.4. The integral! connection between the slab and the beam in cast in-situ construction makes the two act integrally, so that some portion of the slah functions as a flange of the beam. It should be noted that the flattgc is effective only when it is on the compression side, i.e., when the beam is in a 'sagging' mode of flexurc(not 'hogging') with the slah on top. Alternatively, if the beam is 'upturned' (inverted T-beam) and it is subjected to 'hogging' moments (as in a cantilcvcr), the T-beam action is effcctive, as thc flange is under compression. Ideal flanged beam action occurs when the flange dimensions are relatively small while the beam is deep - a s in the case.of closely spaced long-span bridge girders in a T-beam bridge. Thc beam is invariably heavily reinforced in such cases.

BEAM SECTION

Fig. 5.8 Doubly reinforced section design - Example 5.4 Design check To ensure xu 5 x ,,,,

, it suffices to establish p, 2 p, IEq. 4.841

Actual d provided: d = 400 - 30 - 8 - 2812 = 348 mm; d'= 30+16/2 +8= 46 mnl Ford'ld = 46/348 =0.132, f, = 345.8 MPa [Table 4.51. [Alternatively, E,, = 0.0035(1-0.132/0.479) = 0.00253 e.f,, = 345.8 MPa (Table 3.2).1 Actual p, provided: p, = 100x18481(250~348)= 2.124 Actual p, provided: p, = 100x8041 (250x348) = 0.924

r

5.8.1 T r a n s v e r s e Reinforcement in Flange The integral action betwccn the flangc and the web is usually cnsured by the transverse bars in the slab and the stirrups in the beam. In the case of isolated flanged beams (as in spandrel bealns of staircases), the detailing of reinforcement depicted in Fig. 5.9(a) may be adopted. The overhanging -~portions of the slab should be designed . ascantilevers and the r&~lorcetnentprovided accordingly. Adequate transverse reinforcement must be orovided near the too of the flanne. Such reinforcement is usually prescnt in the form of negative rnotncnt reinforcement in the continuous slabs which span across and form the flanges of the T-beams. When this is not the case (as in slabs wltere the main bars run pa,a,ullel to the beam), the Code (CI. 23.1.lb) specifies that transverse reinforcement should be provided in the flange of the T-beam (or Lbeam) as shown in Fig. 5.9(b). The area of such steel should be not less than 60 percent of the main area of steel provided at the midspan of the slab, and should extend on either side of the beam to a distance not less than onefourth of the span of the beam

-

Asp, is slightly less than pf , the section is slightly over-rcinforced. [This can also be verified by applying Eq. 4.81, which gives x,,/d = 0.505 > x , /d = 0.479.1

Revised design To ensure ductile failure,

*

. 9 9 6 ~ 2 5 0 ~ ~=4867 8 ' 4 " "Pc ' ~bd = 0100

iNllz

Provide 3-206 for com~ressionsteel [ A,, = 3x314 = 942 id> 867 - as OK. shown in Fig. 5.81: p, = 100 x 942/(250 x 348) = 1.083 > p,

$

'.

1

' Where the slab and beam are not test monalithically, nvnged beam action cannot

be assumed, unless special s l m r connectors are providcd at the interface bctween beam and the slab.

DESIGN OF BEAMS

204 REINFORCED CONCRETE DESIGN

AND ONE-WAY

SLABS FOR FLEXURE 205

exceeds M,,J,,~, for a singly reinforced flanged section, the depth of the section should be suitably increased: otherwise, a doubly reinforced Section is to b e designed.

Neutral Axis w/thln Flange (xu5 Dl):

Fig. 5.9 Detailing of flanged beams to ensure integral action of slab and beam.

This is, by far, the most common situation encountered in building design. Because of thc very large compressive concrete area contrib11:ed by the flange in T-beams and Lbeams of usual proportions, the neutral axis lies within the flange (4, SD,), whereby the section behaves like a rectangular section having width bland effective depth d. A simple way of first checking x,, < Dj is by verifying M,, < (M,,R)x,,=DI where

(M,,n),=DI is the limiting ultimate moment of resistance for the condition x,, = Dl 5.8.2 Design Procedure In the case of a continuous flanged beam, the negative moment at the face of the support generally cxcecds the maximum positivc moment (at or near the midspan), and hence governs the proportioning of the beam cross-section. In such cases of negative moment, if the slab is located on top of the beam (as is usually the case), the flange is under flexural tension and hence the concrete in the flange is rendered ineffectivc. The beam section at the support is therefore to bc designed as a rectangular section for the factored negative moment' . Towards the midspan of the beam, however, the beambehaves as a proper flanged beam (with the flange under flexural compression). As the width of the web b,, and the overall depth D are already fixed from design considerations at the support, all that remains to be determined is the area of reinforcing steel; the effective width offlange is determined as suggested by the Code [Eq. 4.301. In simply supported flanged beams, howcver, the web dimensions must also b e designed (if not otherwise specified). The width of the web is generally fixed as 250 m n 300 mm, 350 mm (as for a rectangular scctiou), and the overall depth assumed to he approximately s p a d l 3 to spadl6. An appmximate estimate of the area of tension steel A,, can be obtained as follows:

where the lever arm z may be taken approximately as 0.9d or (d- D1/2), whichever is larger. If convenient, the reinforcement should bc accommodated in one layer although, often this may not be possible. When the tctision steel is provided in more than one layer, the effective depth gets reduced. The determination of the actual reinforcenicnt in a flanged beam depends on the location of the neutral axis x,,, which, of course, should be limited to x,,,,,,,,. If M,,

'

In such cnscs it is desirable to distribute the lension steel nt the top of the web 'across the effective width o l the flange, to protect the integral flange from cr~cking- as recommended by the ACI Code. Alternalively,additional reinforcclnent may be provided in the flange region far this porpose.

It may be noted that the above equation is meaningful only if x,,,,,,, > Df In rare situations involving very thick flanges and relatively shallow beams, x,,,,,,, may be in place of D, in less than Df In such cases, M,,J,,, is obtained by substituting x,,,,,,,, Eq. 5.19.

Neutral Axis within Web (xu> Dl): When M,, > ( I W , , ~I,,= ) , it follows that x,, > D,G The accurate determnatton of x,, can be somewhat laborious'. As explained in Chapter 4, the contributions of the compressive forces C,,,, and C,# in the 'web' and 'flange' may b e accounted for sepatately as follows:

M , , = C,,,, t d - O . 4 1 6 ~ , , ) + C , ~ ( d - ~ ~ / 2 )

(5.20)

C,,,,= 0 362fd,,xL,

(5.21)

where,

and the equivalent flange thickness yf is equal to or less than Df depending on whether x,, exceeds 70113 or not. the value of the ultimate moment of resistance For x,,,,,,,27Df/3, (M,,R)r,,=7D,,3 corresponding to x,, = 70,/3'and y l = Dl may be first computed. If

' As an alternative to this procedure, a design based an the appmximate estimate of A,,

[Eq. 5.181 may be assumed, and the resulting section nmlysed to determine M,,* R e design becomes acceptable ifM,,R2 M,, and& < x ,,,,,,-.

206

REINFORCED CONCRETE DESIGN DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE

207

l~isert~ng the appropriate value - D, or the enpresslon for yf (given by Eq. 5 23) - In Eq. 5.20, the resulting quadrat~cequatlon (in terms of the unknown x,,) can be solved to yield the correct value of x,,. Corresponding to this value of x , , the values of C,, and C,$ can be computed [Eq. 5.21, 5.221 and the required A,, obtained by solving the force equ~libriumequatlon.

EXAMPLE 5.5 Design the interior beam in the floor system in Example 5.3 [Fig. 5.4(a)]. Assume that the beam is subjected to moderate exposure conditions. Use Fe 415 steel.

.

SOLUTION

.

The slab is one-way, spanning between the beams, which are simply supported and hence behave as T-beams [ l o = 8230 m , DJ= 160 mm, 6," = 300 m l . Effectiveflange width (CI 23.1.2 Code): 6, = lo /6+b,, +6Df [Eq. 4.301

-

= 30.8 mm

[Alternatively, this is obtainable from Table A.6.1 It may be observed that the bars (either 3-364 or 4-324) can be accommodated in one layer, given 6,"= 300 mm. Assuming 32 mm4 bars and 8 imn4 stirrups, a Actual d = 550-324-3212 = 494 m m (clear cover shall not be less than the diameter of the bar) Deternzirzirrg actual A,, xu,,2,, = 0.479~494= 237 mm > Dl= 160 m . Assuming the neutral axis to be located atx,, =Dl ( M , , , ) , , + = 0 . 3 6 2 ~ 2 5 ~ 2 6 3 2 ~ (494-0.416~160) 160~ [for M 25 concrete]

. .

= 1 6 2 9 x l 0 ~ ~ m>mM,, = 484 kNm Hence, the neutral axis is located definitely within the flange (s,, < Dl). Accordingly, designing the T-scction as a singly reinforced rectangular section with 6 = 6, = 2632 nnli and d = 494 nun,

= 8230/6+300+(6x160) = 2632 Imn,

which is acceptable as it is less than 6, +clear span of slah (300+3400 = 3700). Assume overall depth D = 1 / 15 = 550 m . a effective depth d =500 mm. a effective span 1 = 8.0 + 0.23 m = 8.23 m (less than 8.0 + 0.5 = 8.5 m).

. ,,

Delernrining M. for design Distributed loads from slah (refer Example 5.3): = 5.5 kN/1n'x3.7 m = 20.35 kN/m =4.0kN/m2x3.7 m = 14.8 kN/m Additional dead load due to self weight of web: A,.,, = 25x0.3x(0.55-0.16) = 2.93 kN/m :.Factored load w, = l.Sx(20.35 + 14.8 + 2.93) = 57.12 kN/m.

,,

..

or, providing 4 bars, @,,,

=, Factored

(which incidentally is about 6 percent less than the approximate value calculated earlier). Provide 2-32 $plus 2-28 4 bars [A,, = (2x804) + (2x616) = 2840 mm2 > 2808 nun2]]. The cross-section of the beam, showing the location of bars, is depicted in Fig. 5.10.

moment (maximum at midspan)

M,, = w,,12 18 = 57.12~8.23~/8 = 484 kNm

Deternrining approxinrale A,, Assuming a lever arm z equal to the larger of 0.9d = 450 mm and d - Df12 = 420 mm, i.e., z = 450 mm, Fig. 5.10 T-beam of Example 5.5

208

DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE

REINFORCED CONCRETE DESIGN

209

Checkfor Deflectiorr Control Ignoring the contribution of flanges (conservative) [refcr Section 5.3.21, p, = l0Dx2840 = 1.92; f, = 0 . 5 8 ~ 4 1 5 ~ 2 8 0=238 8 MPa 300x 494 2840 k, = 0.844 [Table 5.21 =, ([Id) ,, =20x0.844~1= 16.88[Eq. 5.51 (lid),,,,,, = 82301494 = 16.66 < 16.88 -Hence. OK.

(a) given section

EXAMPLE 5.6

A continuous T-beam has the cross-sectional dimensions shown in Fig. 5.11(a). The web dimensions havc been determined from the consideration of negative moment at support and shear strength requirements. The span is 10 m i n d thc dcsign moment at midspan under factorcd loads is 800 W m . Determine the flexural reinforcement requirement at midspan. Consider Fe 415 steel. Assume that the beam is subjected to moderate exposure conditions. SOLUTION Detemirrirrg approxirnnfe A,, Effectiveflange width bf Actual flange width provided = 1500 mm; Dl= 100 nmi, b , = 300 mm. Maximum width permitted = (0.7 x 10000)16 + 300 + (6 x 100) = 2067 mm > 1500 mm. .,.b, = 1500 mm Assuming d = 650 mm and a lever arm z equal to the larger of 0.9d = 585 mm and d - Df/2 = 600 mm, i.e., z = 600 mm, 800x106

= 0.87x425x6W = 3693 mm2

Providing 4 bars,

=

$,eqd

- = 34.3 m,i.e,, 36 m.

As 4-364 bars cannot be accommodated in one layer within the width b,, = 300 m,two layers are required. Assumine a reduced d = 625 m, z 625 - 10012 = 575 mm.

-

Provide 5-324 [A,, = 804 x 5 = 4020 mm2] with 3 bars in the lower layer plus 2 bars in the upper layer, with a clear vertical separation of 32 mm - as shown in Fig. 5.1 l(b). Assuming 8 mm stirrups and a clear 32 mm cover to stirrups, 1-,d=700-32-8-

1 5

-[(3~16)+2~(32+32+16)1

Fig. 531 T.beam

(b) proposed reinforcement

of Example 5.6

Determining act!mlA,,

..

x,,Df = 100 mm, the condltlon x. = Dfsatlsfies XU . Expressions for V,,, arc derived in Section 6.7.4. From a design viewpoint, suitable shcar reinforccment has to bc designed if the factored shear V,, exceeds V,,, (i.e., Z, cxceeds lsJ, and the shear rcsistmce required Crom the web minforcemcnt is given by

,

> !J - ! J

= (T" - % = )bd

(6.16)

6.7.3 Limiting Ultimate Shear Resistance

As explained earlier, the yielding of the shear reinforcement at the ultimate limit state is essential to ensure a ductile failure (with ample warning). However, such a failure will not occur if the shear reinforcement provided is cxcessive. If the total crosssectional area A,, of the stirrup Legs and the bent-up bars exceeds a certain limit, it is likely that the section becomes stronger in diagonal tension compared to diagonal compression. Hencc, a ~.hea+cornpression failure [Fig. 6.4(c)] may occur even before the shear reinforcement has yielded (and thus realised its full potential). Such a situation is undesirable due to the brittle nature of the failure: moreover, it turns out to be uneconomical, in much the same way as over-reinforced beams. In order to prevent such shear-compression failures and to ensure yielding of the shear reinforcement at the ultimate limit state, the Code (Cl. 40.2.3) has indirectly imposed a limit on the resistanceV,, by limiting the ultimate shear resistance V,,$

where z ,,, (= 0 . 6 2 G ) is given values (in MPa) of 2.4, 2.8, 3.1, 3.4, 3.7 and 3.9 for concrete grades LM15, M 20, M 25, M 30, M 35 and M 40 respectively (vide Table 20 of the Code). Thus, if the calculated nominal shear stress z,, (= !J,/bd) at a beam section (or thc factored shear Corce V,, exceeds V,,R,M,,~), the design exceeds the limit z, should be suitably revised, either by improving the gratlc of concrete (thereby, raising

z,,,,,,) or increasing the dimensions of the beam (thereby, lowering z,).

245 The

increase in z,,,,, with the compressive strength of concrete follows logically lrom the fact that the shear strength in diagonal compression gets enhanced. In the case of solid slabs, the Code (CI. 40.2.3.1) specifies that r , should not exceed 0.5 T ,,, (i.c., 1.2 MPa for M 15, 1.4 MPa for M 20, 1.55 MPa for M 25, 1.7 MPa for M 30, 1.85 MPa for M 35 and 1.95 MPa for M 4 0 concrete).

6.7.4 Shear Resistance of Web Reinforcement Traditionally, the action of web reinforcement in reinforced concrete beams has been explained with the aid of the truss analogy, the simplest form of which is shown in Fig. 6.9. This design model was first enunciated by Ritter in 1899. In this model, a reinforced concrete bcam with inclined cracks is replaced with a pin-jointed truss, whose compression chord represents tlic concrete compression zone at the top, and whose tension chord at the bottom represents the longitudinal tension reinforcement. Fnrther, the tension web members (shown vertical in Fig. 6.9a) represent the stirrups, and the diagonal web members represent the concrete in compression between the inclined cracks. (The truss model is aki~ito the Strut-and-Tie model). In this model, the compression diagonals do not have to go from top of one stirrup to the bottom of the next. In reality, rather than having discrete diagonal compressive struts, there is a continuous field of diagonal compression contributing to shear resistance. The truss model is a helpful tool in visualising the forces in the stirrups (under tension) and the concrete (under diagonal con~pression),and in providing a basis for simplified design concepts and methods. However, this model does not recognise fully the actual action of the web reinforcement and its effect on the various types of shear transfer mechanisms identified in Fig. 6.3. Fig. 6.9(b) shows one segment of the beam scparated by a diagonal tension crack. This is an idealizationof Fig. 6.9(a), wherein the diagonal crack is assumed to be straight, inclined at an angle 0 to the beam axis and extends over the full depth of the beam. The general case of inclined stirrups is considered in the freebody in Fig. 6.9(b); only the forces in the web reinforcement that contribute to the resistance V, are shown. The inclined stirrups are assumed to be placed at an angle a (not less than 45' in dcsign practice) with the beam axis, and spaced s, apart along the beam axis. If A,, is the total cross-sectional al.ea of one stirrup (considering all the legs intercepting the inclined crack) and 0.87f, is the design yield stress in it (assuming yielding at the ultimate limit state), then the total shear resistance of all the sti11.ups intercepting the crack is given by: V,, = (vertical component of tension per stirrup) x (number of stirrups) =, V,, = (0.87f,A,, sin a ) x d(cot 0 +cot a)/s,.

DESIGN

246 REINFORCED CONCRETE DESIGN

FOR SHEAR

247

b a s are bent-up at the sanie location at an anglea , the vertical component of the total tension in thesc bars is given' by

STIRRUPS 7

6.7.5 Influence of S h e a r o n Longitudinal Reinforcement I

The truss analogy illustrates an inlportant effect pcrtaining lo influence of shear on the tension in the longitudinal reinforcement. Usually, the tension steel area requirement at any section is governed by,,the bending moment in the beam at that section. However, when the beam is cracked (especially, at ultimate loads), there will b e a change in the calculated tensile stress. The presence of a diagonal crack will alter the tensile stress in the longitudinal steel, as observed earlier in the context of curtailment of bars (refer Fig. 5.14). This is also clear in the truss analogy, as revealed by the section of the truss shown in Fig. 6.9(c). By applying the 'method of sections', wc observe that the compressive force in the top chord will be less than thc tensilc fol-cc in the bottom chord of the truss in any given panel (owing to the prescnce of the diagonals) and this diffc~wcewill be equal to the horizontal component of the force in the diagonal. Whcreas the force in the top chonl (compression in concrete) is governed by the bcnding momcnt a1 A, the force in the bottom cho1.d (tension reinforcement) is governed by the bending moment at B, which is higher than that at A. Thus, the presence of a diagonal tension crack due to shear results in an increase in the tension in the longitudinal reinforcement. The increased tension is given approximately by half of thc horizontal component of the force in the diagonal strut in Fig. 6.9(c)'; i.e., equal to 0,5V/ta,iB. This influence of shear in enhancing the longitudinal reinforcement rcqnirement was not realised till the 1950s. Even now, this is not directly rcflected in the IS. Code provisions as a specified additional

BARS

(c) effect of shear on longitudinal reinforcement

. ..

Fig. 6.9 Classical truss analogy for action of web reinforcement Assuming, for convenience, that the crack is located at 8 = 45', the above relation simplifies to (6.18) V,,, = 0.87 f,A,,, (d/s,)(sina + c o s a ) The case of 'vertical stirrups' may he considered as a special case with a = 90'. Hence, for vertical stirrups, thc shear resistancc V , ,is obtained from Bq. 6.18 as

V,,, =0.87 f,A,,d/s,

Sections 5.9.2 and 5.9.3). The Code (CI. 26.2.3.1) requires that the flexural tension reinforcement be extended for a distance of d or 12$, whichever is greater, beyond the location required for flexure alone. Here 4 is the nominal diameter of the longitudinal bar concerned, and the provision is applicable for locations other than at the supports of simple spans and at the free ends of cantilevers with concentrated loads. This provision is equivalent to the outward shifting of the design moment diagram by a distance of d or 12$ (Fig. 6.10a).

(6.19)

The she= resistance of bent-up b a n may also be obtained from Bq. 6.18 -when a series of single or parallcl bent-up bars are provided at regular intervals in the nlanner of inclined stirrups. However, when a single bar or a single group of parallcl

'It may be noted that Eq. 6.20 is applicable only in the limited iegior where the bar is bent up. 'The horizontal component of the force in the diagonal sIr.lt, equal to V/tmB is assumed to be balanced equally by forces in tho tap and bonom chords. This, incideatally, also implies that there will be a reduction in the longitodinal con~pressionin the top chord, equal to 0.5 V/mrtB.

248 REINFORCED CONCRETE DESIGN

-1

dor

12 du

,

DESIGN FOR SHEAR 249 that every potential diagonal crack is intercepted by at least one stirrup. Further, the Code specifies that "in no case shall rhe spacing exceed 300 m d . The overall shear resistance V(,Ris given by Eq. 6.14. For the purpose of design for a given factored shear force V,,,the web reinforcement is to be designed for a design shear force of (V,, - z,bd ), provided z, S z ,", (i.e., V,,< V,,R,,i,,,), From the viewpoint of analysis of a given reinforced concrete bcam, the capaciiy V , may be determined from Eq. 6.18 - 6.20, assuming that the steel has yielded. However, the total shear resistance V,,,q,given by Eq. 6.14, should be limited to V,,R,li,,, given by Eq. 6.17.

tension steel. including shear effect

(a)

(b)

Fig. 6.10 Design bending moment for tension steel, including shear effect

6.7.6 Minimum Stirrup Reinforcement The Code (C1.26.5.1.6) specifies a minimum shear reinforcement to h e provided in the form of stirrups in all beams where the calculated nominal shear stress Z, exceeds 0 . 5 : ~ ~

At simple supports and near free ends of cantilevers, the flexural tension reinforcement should be capable of resisting a tensile force of Vt - 0.5 V, at the inside edge of the bearing area, where V, is the factored shear resistance provided by the shear reinforcement in this location. This expression can be derived from equilibrium considerations of the forces in the free body diagram of the support region, separated by a diagonal crack [Fig. 6.10bI. Taking moments about A and neglecting small quantities of second order,

>L when zu> 0.5 z, -

bs,

(6.22)

0.87 f,

The maximum spacing of stirrups should also comply with the requirements described earlier. For normal 'vertical' stirrups, the requirement is

V,Z = Tz+V, 212

If the actual straight embedment length available at the support, x, is less than the development lcngth, 4 , the stress that can be developed in the bar at the critical section at the inside edge of the bearing area (Fig. 6.10b) may be taken as: fs = $ s f y ( ~ / ~ d )

Alternatively, the embedment length required to develop the stressf, in the bar can be computed as:

The Code objective in recommending such minimum shear reinforcement is to prevent the sudden formation of an inclined crack in an unreinforced (or very lightly reinforced) web, possibly leading to an abrupt failure. Further, the provision of nominal web reinforcement restrains the growth of inclined shear cracks, improves the dowel action of the longitudinal tension bars, introduces ductility in shear and provides a warning of the impending failure.

6.6 ADDITIONAL COMMENTS ON SHEAR REINFORCEMENT DESIGN

Code Recommendations Eq. 6.18-6.20 are given in the Code under Ci. 40.4. Further, the Code limits the maximum value off, to 415 MPa, as higher strength reinforcement may he rendered brittle at the sharp bends of the web reinforcement; also, a shear compression failure could precede the yielding of the high strength steel. The Code (C1.26.5.1.5) also limits the value of the spacing s,. to 0.75 d for 'vertical' stirrups and cl for inclined stirrups with a = 45'. This is done to ensure

Bent-up bars generally give lower shear strength and often result in wider cracks than stirrups. Hence, unless there is a series of such bars bent up at relatively close spacings (as is possible in long-span bridge girders), there is not much economy resulting from considering their shear strength contribution. In normal situations, where there are only a few isolated bent-up bars scattered widely along the span, their shear strength contribution (not available at all sections) is ignored. Accordingly, the stirrups are designed to carry the full excess shear, given by:

250

. . e

~EINFORCED CONCRETE

i

DESIGN

DESIGN FOR

'hanger bar' of nominal diameter) must be located at every bend' in a s t i ~ ~ u T p .h e ends of the stirrup enclosing the longitudinal bars should satisfy anchorage requimnents (discussed in Chapter 8). Although the Code does not call fo; shear reinforcement in pot.tions of beams where 2, < < , / 2 , it is good design practice to provide minimum (nominal) stirmps [Eq. 6.231 in this region - to improve ductility and to restrain inclined cracks in the event of accidcntal overloading. The factored shear force V1 to be considered for design at any section must take into account possible variations in the arrangement of live loads. The construction of shear envelope for this purpose is demonstrated in Examples 6.1 and 6.3. Termination of flexural reinforcement in the tension zone can lower the shear strength of bcams (refcr Section 5.9). Hence, such scctions may also be critical and have to be checkcd for shear; if necessary, additional stirrups should h c provided ovcr a distance of 0.75d from the cut-off point to satisfy the Code requirement (CI. 26.2.3.2). This is demonstrated in Examples 6.1 and 6.3. When reversal of stresses occurs, as in the case of earthquake loading or reversed wind direction, the shcar strength of the (previously cracked) concrete cannot b e relied upon. In such cases, the stirrups should be designed to take the entire shear. Moreover, the stirrups should necessarily bc in the form of closed loops placed perpendicular to the member axis. [The details of earthquake-resistant design for shear are described in Chapter 14.1

Inclined stinups are most effective in reducing the width of the inclined cracks, and are desirable when full depth transverse cracks are likely (as in beams with high axial tension). However, such reinforcement may be rendered entirely iiieffective if the direction of the shear force is reversed (as under seismic loadst ). 'Vertical' stirrups are the ones most commonly employed in practke. It should be noted that the use of closely spaced stirrups of smaller diameter gives better crack control than stirrups of larger diameter placed relatively far apart. The diameter is usually 8 mm,10 mm or 12 tmn. Where heavy shear reinforcement is called for, multiple-legged stirrups should be employed (as often required in the beams of slab-beam footings). For n-legged stirrups of diameter @, (where n = 2 , 4 , 6), A,, = nm$j 14

(6.26)

The required spacing s, of 'vertical stirrups' for a selected diameter @, is given by applying Eq. 6.19, as: 0.87fYA,, (6.27) s, 5 ",Jd where (from Eq. 6.251,

6.9

It can also be seen from Eq. 6.19 and Eq. 6.26, that for a given amangement of vertical stit~ups(with specified n, $I,, s,), the shear resistance in terms of V,Jd is a constant (in N/mm units) given by V,,, 0.87fy4, -= (6.29)

.

SHEAR 251

8

Accordingly, suitable design aids can be prepared expressing the above equation, as done in Table 62 of SP : 16 [Ref. 6.91 - to enable a quick design of vertical stirrups, for a specified V,,ld. The stirrup bar diameter is usually kept the same for the entire span of the beam. Theoretically, the required spacing of stirrups will vary continuously along the length of the beam owing to the variation in the shear force V,,. However, stin'ups are usually arranged with the spacing kept uniform over portions of the span satisfying the requirements of she= strength [Eq. 6.271 and maximum spacing [Eq. 6.23, 6.241. The first stirrup should be placed at not more than one-half ) the face of the support*. Also, a longitudinal bar (at least a spacing ( ~ " 1 2 from

'Special prnvisians for shear reinforcement design, under earthquake loading, are covered in Chapter 16. 'See Section 6.5 and Fig. 6.7 regarding special case involving shear hansfer.

INTERFACE SHEAR AND SHEAR FRICTION

6.9.1 Shear-Friction

There are situations where shear has to be transferred across a defined plane of weakness, nearly parallel to the shear force and along which slip could occur (Fig. 6.11). Examples are planes of cxisting or potential cracks, interface between dissimilar materials, interfaces between elements such as webs and flanges, and interface between concrete placed at different times. In such cases, possible failure involves sliding alang the plane of weakness rather than diagonal tension. Therefore it would be appropriate to consider sheat resistance developed along such planes in the form of resistance to the tendency to slip. The shear-jriction concept is a method to do this. When two bodies are in contact with a normal reaction, R, across the surface of contact, the frictional resistance, F, acting tangential to this surface and resisting where p is the coefficicnt of friction (Fig. 6.12a). relative slip is knoyn to be F = Figure 6.12(b) shows an idealised cracked concrete specimen loaded in shear. In such a specimen, a clamping forcc between the two faces of the crack can be induced by providing reinforcement (shear-friction reinforcement, A,,) perpendicular to the crack surface. Any slip between the two faces of the rough irregular crack causes the

m,

!

'The stin.up is tied to the longitudinal bar using 'binding wire'

faces to ride upon each other, which opens up thc crack (Fig. 6.12~). This in turn induces tensile forces in the reinforccment, which ultimately yields (Fig. 6.12d). If the arca of reinforcement is A,, and yield strcss f , , at ultimate, the clamping force between the two faccs is R = A& and the frictional rcsistance is I: F = A,,fp.. (a)

Fictional farce

A",

patentia crac

I (a) Corbel

crack

... (b) Precast beam seat

I (c) Column face plate (b)

Fig. 6.11 Typical cases where.shear friction is applicable (adapted from Ref. 6.10)

In reality, the actual resistance to shear, V,, is conlposed of this frictional force (2 0, the resistance to shearing off of the protrusions on the irregular surface of the crack, the dowcl force developed in the transverse reinforcement, and when there are no cracks developed yet, the cohesion between the two parts as well. The nominal or cha,octel.istic ( i e , without safety factors) shear resistance, V,,,, due to the friction betwcen the crack faccs, is given by EL].6.30. Other less simple methods of calculation have been proposed (Rcfs. 6.11, 6.12) which result in predictions of shear transfer rcsistance in substantial agreement with comprehensive test results. For shear-friction reinforcement placed pc~pendicularto the shear plane,

vn = A,f# (6.30) nominal shear resistance due to the assumed friction part alone contributed by reinforcement stress Avf = area of shear-friction reinforcement, placed normal to the plane of possible slip fi = coefficient of friction. Shear-friction reinforcement may also be placed at an angle q t o the shear plane, such that the shear force produces tension in the shcar-friction reinforcement, as shown in Fig. 6.13(a), (b) (i.e., q i; .!s x..

?J g. s'.:

f

h

S O I I I ~ (rdaively

rare) siuxiom, a x i d form (tension or mnprcssioto m y a l a be

clearly understood that this is ,nelcly a matter of terminology, and that it does not equilibriuul conditions need not be satisfied in cases of 'compatibility

268

REINFORCED CONCRETE DESIGN

DESIGN FOR TORSION

There are some situations (such as cifcular beams suppotted on multiple columns) where both equilibrilrrn tonion and compntibiliry torsion coexist. 7.2.1 Equilibrium Torsion This is associated with twisting moments that are developed in a structural inembcr to maintain static equilibrium with the external loads, and are independent of the to 0 (i.e., M, > M.), then a reinforcement area A,,' is to be designed to resist this equivalent moment, and this steel is to be located in the 'flexural compression zone'. It follows from the above that in the limiting case of 'pure torsion' (i.e., with M,,=O), equal longitudinal reinforcement is required at the top and bottom of the rectangular beam, each capable of resisting an equivalent bending moment equal to

w.

7.4.4 Design Strength in Torsion Combined with Shear Tonion-shear interaction curves have been proposed [Ref. 7.161, similar to torsionflexure interaction curves. In general, the interaction between T,,/~;,R and V,/K,n takes the following form:

T,,and V,, are the given fact&ed twisting moment and factored sheax force

r&pectivel'y; zr and VLIRare the ultimate strengths in 'pure torsion' and 'flexural shear' (without torsion) respectively; a is a constant, for which values in the range 1 to 2 have been proposed [Fig. 7.91. A value of a equal to unity results in a linear interaction and generally provides a conservative estimate.

The Code provisions (CI. 41.4.3) for the design of transverse stiuup reinforcement (2-legged, closed) are bascd on ihc skew-bcnding thcoty and are aimed at resisting a Mode 2 failure [Fig. 7.7(b)j, causcd by a large torsion combined with a small flcxnral shear:

where A,, = 24, is the total nrca of two legs of the stirrup: s,. is tile cent^-to-centre spacing of the stirrups; b l and dl are the centre-to-centre distances between the comer bars along the width and depth respectively; and T,, and V,, are the factored tu isting nionic~~l alxd i a m w 4 4w.tr im.c :cling LLIlhc .,cciiun w.I:r c u 1 ~ 4 c r ~ l i ~ n . It may he obscrvcd 1l1aI iol 11,. r ~ h ~ ~ ccni i~c\ eoiitlc111?1Iii n ' I ~ I I I Ctotw11' 11 . c... with XI = 0 and 7;, = T,,& Eq. 7.23 becomes exactly equivalent to Eq. 7.14, which was derived using the space-truss analogy. In addition to Eq. 7.23, the Code (C1. 41.4.3) specifies a nunimuni limit to the toral area of transverse reinforcement: ~

where z, is the 'equivalent no~iunalshear stress' give11 by Eq. 7.7. The purpose of Eq. 7.24 is to provide adequate resistance against flexural shear failure, which is indicated in situations where T , is negligible in cornpalison with V,. Indeed, for the extreme case of T,= 0, Eq. 7.24 becomes exactly equivalent to Eq. 6.25, which was derived for flexural shear. It may benoted that the contribution of inclined s t i ~ ~ u p s and bent up bars can be included in the calculation of A,, in Eq. 7.24, but not Eq. 7.23.

Distribution of Torsional Reinforcement

This formula can alternatively be generated from the space tlvss analogy [Fig. 7.6(a)l, by visualising the longitudinal tensile forces in the bars (located either at top or at bottom) as those required to resist an effective bending moment M, which can be shown to be equal to T,,(I+ d,&J; the Code has simplified this formula to the form given in Eq. 7.20.

The Code (CI. 26.5.1.7a) specifics maximum limits to the spacing s, of the stirrups provided as torsional reinforcement - to ensure the developmnent of post-cracking

DESIGN FOR TORSION

284 REINFORCED CONCRETE DESIGN

torsional resistance, to control crack-widths and to control the fall in torsional stiffness on account of torsional cracks: s,

s

(7.25)

(XI+ Y I ) / ~ mm

285

EXAMPLE 7.2

The beam of Example 7.1 is reinforced (using Fe415 grade steel) as shown in Fig. 7.10(a). Determine the design torsional resistance of the beam under pure torslon. Assume moderate exposure condition. SOLUTION

where xl and yl are, respectively, thc short and long centre-to-centre dimensions of & should satisfy all the limits given in the rectanzular closed stimps. The spacing . Eq. 7.25. The Code (C1. 26.5.1.7b) also recommends that the "longitudinal reinforce~nent shall be placed a s close a s is practicable to the comers of the cross-section, arid in all cases, there shall be nr least orre longitudinal bar in each corner of rhe ties". Further, if the torsional member has a cross-sectional dimension (usually, overall depth rather than width) that exceeds 450mm, additional longitudinal bars are required to be provided as side face reinforcement, with an area not less than 0.1 percent of the web area. These bars are to be distributed equally on the two faces at a spacing not exceeding 300 mm or web thickness, whichever is less.

-

~

7.5 ANALYSIS AND DESIGN EXAMPLES F I ~7.10 . Example 7.2

EXAMPLE 7.1

A plain concrete beam (M 20 grade concrete) has a rectangular section. 300 mm wide and 500mm deep (overall). Estimate the 'cracking torque'. Also determine the limiting torque beyond which torsional reinforcement is required (as per the Code), assuming T, = 0.3 MPa.

.

Using theplastic Nteory forr~rrrla[Eq. 7.51: 1 T,, = -z 2 ,,,,bz(~- b/3) where b = 300 mm, D = 500 mm. Assuming 7,,,,,,, = 0 . 2 6 = 0 . 2 m = 0.894 MPa.

= 16.09 x lo6 Nmm = 16.1 kNm. As per IS Code formulation, torsion has to be combined with shear for deciding whether or not torsional reinforcement is required. Torsional reinforcement is required if z, >T, , i.e.,

V,, + 1.6 T,, /b

bd Assuming V,,= 0, d = 0.9D = 450mm and z, = 0.3 MPa,

'Tc

Given b = 300 mm, D = 500 mm,hk = 20 MPa,f, = f9 = 415 MPa, Al(dueto4-16 $ p l u s 2 - 1 0 ~ ) = ( 2 0 1 x 4 ) + ( 7 8 . 5 x 2 ) = 9 6 1 m 2 . A, ( l o $ stimp) = 78.5 mm2,A;=2A, = 157 mm2. s, = 150 mm b1=300-30x2-10x2-16=2041mn dl = 500 - 30 X 2 - 10 x 2 - 16 = 404 mm [Fig. 7.10(b)]. Applying the general space truss fonnulation, considering the contribution of both transverse and longitudinal reinforcements [Eq. 7.181:

= 38.27 x lo6Nmm = 38.3 kNm

..

.,,

which is greater than T, = 16.1 IcNm (refer Example 7.1). Alternatively, using the IS Code formula, considenng shear-tors~on mteraction [Eq. 7.231 with V. = 0, which corresponds to the space truss fonnulat~on considering the contribution of the transverse remforcement alone [Eq 7.141: T,R= A & ~ I (0.87fy)/s,

= 157 x 204 x 404 x (0.87X415)&0 =31.1 x 1 0 ~ ~ m m = 3 lk. l~ m .

286

REINFORCED CONCRETE DESIGN

DESIGN FOR TORSION

Assuming a load factor of 1.5, Factored distributed load w,,= 8.75 x 1.5 = 13.13 kN/m. Eccentlicity of cantilever load from bcam centreline = 1.012+ 0.312 = 0.65 m. .'.Factored distributed torque f,, = (5.0 x 0.65) X 1.5 = 4.88 kNm/~n Stress resultants:

, ' , In

the above formulation, we had tacitly assumed that the torsional strength is governed by shear considerations and not 'equivalent moment'. The reader may Yerify this assumption by checking the equivalent moment capacity due to the longitudinal reinforcement' using Eq.7.20.

EXAMPLE 7.3

Max. twisting moment (at suppor!) ?;, =

A beam, framing between columns, has an effective span of 5.0 m and supports a cantilevered projection. 1 m wide [Fig. 7.11(a)] throughout its length. Assume that the cross-sectional details of the beam are exactly the same as in :Example7.2 [Fig. 7.10l. Determine the adequacy of the section (as per IS Code), assuming a total uniformly distributed load (DL+LL) of 5 kNlm2 on the cantilever pro~ection as shown. Assume fixity at the ends of the beam against torsion as well as flexure.

Max, bending moment (at support) M,,

= Aw

lZ

12

= 27.35 kNm (hogging).

r,,

Need for torsioual reinforccrncnt Equivalent nornimd slterrr swess [Eq. 7.91: V, 20 mm OK.

REVIEW QUESTIONS

Fig. 7.12 Example 7.4

.

Design of transverse reinforcement Eouivalent nominal shear stress:

=3,06MPa - (110x10~)+1.6~(140x10~)/3~0 350x700 c T,,,,

=3.1 MPa (for M 25 concrete)

-

r

Shear strength of concrete .IEa. . 6.101:

e

For p, = 2214x100 = 0.904, = 3.21 1 =, r , = 0.618 MPa (for M 25 concrete) 350 x 700 As torsional shear is relatively high, Eq. 7.23 is likely to goveln the design of stin.ops (rather than Eq. 7.24). Assuming 10 4 2-legged stirmps, A,, = 78.5 X 2 = 157 mm2.

P

... . .., With 50 mm effective cover assumed all around, [Fig. 7.121. dl = 7 5 0 - 5 O x 2 = 6 5 0 m m

Explain, with examples, the difference between equilibrium tomion and compatibiliry torsio,t. "Equilibrium torsion is associated with statically deterl~natc structures, whereas compatibility torsion is ass~ciated with statically indeterminate structures". Is this statement true? Comment. Reinforced concretc colunms are rarely subjected to torsion. Cite an cxample whcre this situation occurs, i.e., torsion exists in combination with axial compression, and perhaps also with flexure and shear. How is torsional st@zess estimated for 'compatibility torsion'? (a) Estimate the torsional stiffness of a reinforced concrete beam element (of a frame), having a span 1 = 6.0 m and a rectangular section with width b = 200 nun and overall depth D = 500 mm. Assume M 25 concrete. (b) Compare the torsional stiffness with the flexural stiffness. 4EI/1, for the same beam element. In the case of a cirrular shaft subject to pure torsion, it i s well known that the maximum torsional she% stress occurs at locations of maximum radius. If the member has a rrcro~zgulur(instead of circular) cross-section, the comer points are the ones located furthest from the shaft axis. However. the torsional shear stress at these points is not the maximum; the stress, in fact, is zem! [refer Fig. 7.3(b)]. Why? ~ k c u s sthe torq&-twist relationship for (a) plain concrere and (b) reirrforced concrere me~nberssubjected to pure torsion.

292

7.7 7.8

7.9 7.10 7.11

REINFORCED CONCRETE

DESIGN

DESIGN

Inclined stirrups and bent-up bars arc considered suitable for shear reinforcement, but not torsional rcinforccment. Why? For a thin-walled tubular section of arbitrary shape (but uniform thickness) subjected to pure torsion T,,,the shearjlow q can be assunled to be constant (in .the plastified state) at all points on thc cenlrcline of the thickness. Using this concept, derive the relationship [Eq. 7.91 between q and T,,in terms of Ao, the area enclosed by the ccntreline of thc.thickness. Briefly explain the concept underlying the space truss analogy for estimating torsional strength of a reinfo~cedconcrete beam. Briefly discuss the different modes of failure under combined flexure and torsion. Bliefly dischss torsion-shear interaction of reinforced concrete beams.

7.5

FOR

TORSION 293

Repeat the Problem7.4, considering an ultimate twisting moment of 5 0 kNm (instead of 25 !dim).

PROBLEMS 7.1

Determine the design torsional resistance of the beam shown in Fig. 7.13 under pure torsion by (i) IS Code proccdure. (ii) general space truss formulation. Assmne M 25 concrete and Fe 415 steel. Arcs. (i) 48.6 kNm (ii) 62.2 kNm

(a) plan : &lar

gird&

300

A

B

+ clear cover = 5 rnm on all sides

TWlSTlNQ MOMENT

(b) stress resultanls at angle B

(c) Variation of

v , Muand r,

Flg. 7.14 Problem 7.6

7.6 ~ l g7.13 . Problems 7.1,7.2 7.2

7.3

7.4

For the beam section shown in Fig. 7.13, dcrive and hence plot suitable interaction relationships (satisfying IS Code requirements) between: (a) torsion and bending (T,,- M,,) (b) torsion and shear (T.- V,,) Plot. T,,on the y - axis in both cases. Consider the problem described in Example 7.3 as a design problem instead of an analysis problem. Consider a total distributed service load of 10 kNlm2 (instead of 5 liii/m2) on the 1 m wide cantilever projection [Fig. 7.1 I]. Design the reinfolcement in the beam section (300 mm X 500 mm), assuming M 25 concrete, moderatc exposure conditions and Fc 415 steel. Design a rectangular beam section, 300 mm wide and 550 mm deep (overall), subjected to an ultimate twisting moment of 25 W m , combined with an ultimate bending moment of 60 W m and an ultimate shear fane of 50 W. Assume M 20 concrete, moderatc exposure conditions and Fe 415 steel.

Consider a circular girder of radius R = 5 m with rectangular cross-section, 350mm wide and 750mm deep (overall), supported symmetrically on 8 pillars [Fig. 7.14(a)l. Design and detail one typical span (AB) of the girder, assuming M 2 5 concrete and Fe 415 steel. The total ultimate (uniformly distributed) load on span AB may be taken as W,, = 1400 !di,inclusive of selfweight of the girder. Expressions for the bending moment (M,,), twisting moment (T,) and shear force (V,,) at any location 8 (in radians) [Fig. 7.14(b)] can be derived from first principles. For convenience, these expressions are summarised below: M,, = (W.R)[OSsinB + 1.2071 1cosB - 1.273241 I;, = ( ~ ~ ) [ 1 . 2 101sinB 7 - 0.5~0~6'-4 8 j n + 0.51

v,,= 0.5 W"(1 - 88/7r) Assume moderate exposure conditions.

284

REINFORCED CONCRETE DESIGN

REFERENCES Timoshenko, S., and Goodier, J.N., Theory of Elasticify, Third edition, McGraw-Hill, New York, 1970. - Explanatoq Handbook on Indian Standard Code of Practice for Plain and Reinforced Concrete (IS 456:1978), Special Publication SP:24, Bureau of Indian Standards, New Delhi, 1983. Popov, E.P., Mechanics of Materials, Second edition, Prentice-Hall, Englewood Cliffs, New Jersey, 1978. Hsu, T.T.C., Plain Concrete Rectangular Sections, Torsion of Structural Concrete, ACI Publication SP - 18, Am. Conc. Inst., Detroit, 1968, pp 207238. Collins, M.P., The Torque-Twist Characteristics of Reirforced Concrete Beams, Inelasticity and Nonlinearity in Structural Concrete. SM Study No. 8, University of Waterloo P m s , Waterloo, 1972, pp 21 1-232. Cowan, H.J., Reinforced and Prestressed Concrete it1 Toniorr, Edward Arnold Ltd., London, 1965. Hsu, T.T.C., Behuviour of Reinfo~ced Concrete Rectangular Menibem, ACI Publication SP - 18, 'Torsion of Structural Concrete' Am. Conc. Inst., Detroit, 1968, pp 261-306. S t r u c t u r a l Use of Concrete :P a t 2 :Code of Practice for Special Circun~stances.BS 8110 : Part 2 : 1985, British Standards Institution, 1985. Hsu, T.T.C., Ultimate Torque of 'Reinforced Rectangular Beanis, ASCE Journal, Struct. Div., Vol. 94, Feb. 1968, pp 485-510. Lampert. P. and Collins, M.P., Torsion. Bending and Confusion -An Attenipr to Establish the Facts, ACI Joumal, Vol. 69, Aug. 1972, pp 500-504. Mitchell, D. and Collins, M.P., Diagonal Compression Field Theory - A Rntmnnl Method for Structural Concrete in Pure Torsion, ACI Journal, Vol. .....~71, Aug. 1974, pi396-408. Collins, M.P., Walsh, P.F., Archer, F.E, and Hall AS., Ultirnate Strength of Reinforced Concrete Beams Subjected to Combined Torsion and Bending, ~ ~ f p u h l i c a t i oSP n - 18, '~orsionof Stmctural Concrete', Am. Conc. Inst., March 1966. Warner, R.F., Rangan, B.V. and Hall, AS., Reinforced Concrete, Pitman, Australia, 1976. Pumshothaman, P.. Reinforced Concrete S t r u c t u ~ Elenrenfs l - Behaviou,; Analysis and Design, Tata McGraw-Hill Publ. Co. Ltd., New Delhi, 1984. Iyengar, K.T.S. and Ram Parkash, N., Recommendation for the Design of Reinforced Concrete Beams for Torsion. Bending and Sheat.. Bridge and Structural Engineer. March 1974. Mattock, A.H., How to Design for Torsion, ACI Publication SP - 18, 'Torsion of Stiuctural Concrete', Am. Conc. Inst., March 1968. ~~~

~

8.1 INTRODUCTION

'Bond' in reinforced concrcte refers lo thc adhesion bctween the reinforcing stecl and the surrounding c o n c ~ t c .It is this hond which is resporlsible for the transfer of axial force from a reinforcing bar to the surrounding concrete, thereby providing stmbz costpati6ility and 'composite action' of concrete and steel [refer Section 1.2.21. If this bond is inadequate, 'slipping' of the reinforcing bar will occur, destroying full 'composile action'. Hence, the lundnmcmal assumption of the theory of flexure, viz. plane sections remain plane cvcii after bending, becomes valid in reinforced concrete only if the mechanism of bond is fully effective. It is th'ough the action 01 bolld resistance that the axial stress (tensile or compressive) in a reinforcing bar can undergo variatioii from point to point along its length. This is required to accommodate the variation in bending moment along the length of the flexural member. Had the bond beell absent, the stress at all points on a straight bar would be constant', as io a string or a straight cable. 8.1.1 M e c h a n i s m s of Bond R e s i s t a n c e

Bond resistance in ~einlorcedco~~cictc is achieved through the following mechanisms: 1. Chemical adhesion - doc to a gunl-like property in the products of hydration (lomed during lhe making of co~~crcte). 2. Frictional resistance - due to thc surface roughness of the reinforcemeiit and the grip exertcd by the concrete shrinkage. 3. Mechanical interlock - due to the surface protrusions or 'ribs' (oriented transversely to thc bat. axis) providcd in deformcd bars.

'

Such a situation is encountered in presrltssed concrete - in anbonded post-tensioned members.

DESIGNFOR

296 REINFORCED CONCRETE DESIGN

Evidently, the resistance due to 'mechanical interlock' (which is considerable) is not available when plain bars arc used. For this reason, many foleign codes prohibit the use of plain bars in reinforced concrete - cxcept for lateral spirals, and for stirrups and ties smaller than 10 mm in diamcter. However, there is no soch restriction, as yet, in the IS Codc.

8.1.2' Bond Stress Bond resistance is achieved by the development or tangential (shear) stress components along the interface (contact surface) between the reinforcing bar and the surrounding concrete. The stress so developed at the interfnce is called bond stress, and is expressed in terms of thc tangential force per unit nominal surface area of the reinforcing bar.

8.1.3 Two Types of Bond There are two typcs of loading situations which induce bond stresses, and accordingly 'bond' is characterised as: I . Flexural bond: 2. Anchorage bond or development bond.

BOND

297

'Flexural bond' is that which a~isesin flexural members on a c c o u ~ of ~ t shear or a variation in bending moment, which in turn causes a variation i n axial tension along the length of a reinforcing bar [Fig. 8.l(d)l. Evidently, flexural bond is critical at points where the shear (V = dM/dx) is significant. 'Anchorage bond' (or 'deQelopment bond') is that which arises over the length of anchorage provided for a bar or near the end (or cut-off point) of a reinforcing bar: this bond resists the 'pulling out' of the bar if it is in tension [Fig. 8.l(e)], or conversely, the 'pushing in' of the bar if it is in compression. These two types of bond are discussed in detail in the sections to follow.

8.2 FLEXURAL BOND As mentioned earlier, variation in tension along the length of a reinforcing bar, o h g to varying bending moment, is made possible through flexural bond The flexural stresses at two adjacent sections of a beam, dx apart, subjected to a differential momcnt dM, is depicted in Pig. 8.l(b). With the usual assumptions made in flexural design, the differential tension d T i n the tensionsteel over the length dx is given by

dT=-

dM

-

(8.1)

L

where z is the lever arm. This unbalanced bar force is transferred to the surrounding concrete by means of 'flexural bond' developed along the interface. Assuming the flexural (local) bond stress u~ to be uniformly distributed over the interface in the elemental length dx, equilibrium of foxes gives:

perimeter = Za

where Zo is the total perimeter of the bars, at the beam section under consideration [Pig. 8.l(c)]. From Eq. 8.2, it is evident that the bond stress is directly proportional to the change in the bar force. Combining Eq. 8.2 with Eq. 8.1, the following expression for the local bond stress u, is obtained:

Alternatively, in terms of the transverse shear force at the section V = dM/dx, T,

dX A=

rt?--T+dT

e

.-,,,, D

B

-

L

*

C

4

u,

US

(d) flexural bond

,

(e)

development bond

~ l g8.1 . Bond stress in a beam

It follows that flexural bond stress is high at locations of high shear, and that this bond stress can be cffectively reduccd by providing an increased number of bars of smaller diameter bars (to give the same equivalent A,,). It may be noted that the actual bond stress will bc influenced byflexural cracking, local slip, splirting and other secondary effects - which are not accounted for in Eq. 8.3. In particular, flexural cracking has a major influence in governing the magnitude and distribution of local bond stresses.

298

REINFORCED CONCRETE

DESIGN

DESIGN FOR BOND 299

8.2.1 Effect of Flexural Cracklng on Flexural Bond Stress From Eq. 8.3(b), it appears that the flexural (local) bond stress ul has a variation that is similar to and governed by the variation of the transverse shear force V. In fact, it would appear that in regions of constant moment, where shear is zero, there would be n o bond stress developed at all. However, this is not true. The tensile force T i n the reinforcement varies between flexural crack locations, even in regions of constant moment, as indicated in Fig. 8.2. At the flexural crack location, the tension is carried by the reinforcement alone, whereas in between the cracks, concrete canies some tension and thereby partially relieves the tension in the steel bars. As local bond stress is proportional to the rate of change of bar force [Eq. 8.21;~ocal bond stresses d o develop in such situations. ,concrete

C

(a) constatlt moment

bond force between the cracks will, of course, be zero in a region oCconstant nloinent. When the moment varies between the flexural cracks, the bond stress distribution will differ from that shown in Fig. 8.2(c), such that the net bond force is equal to the unbalanced tension in the bass betwcen the cracks [Eq. 8,2]. Beam tests show that longitudinal splitting cracks (end to get initiated near the flexural crack locations where tthc local peak bond stresses can be high. Thc use of large diameter bars particularly rcnders the beam vulnerable to splitting andlor local slim Finally, it may be noted that flexural cracks are generally not present in the compression zone. For this reason, flexural bond is less critical in a compression bar, compared to a tension bar with an identical axial force.

8.3 ANCHORAGE (DEVELOPMENT) BOND As mentioned earlier, anchoqrgc bond or development bond is the bond developed near the extreme end (or cut-off point) of a bar subjected to tension (or compression). This situation is depicted in the cantilever beam of Fig. 8.3, wherc it is seen that the tensile stress in the bar segmcnt varies from a maximum (f,) at the continuous end I) to practically zero at t l ~ discontinuous c end C.

region between flexural cracks

(b) probable

variation of bar tension T

~

, , variation of flexural (local) bond stress ut

_L_ly--l

(a) cantilever beam

(b) probable variation

(C)

of anchorage bond stress ua

assumed uniform average bond stress u,,

Flg. 8.3 Anchorage bond stress Fig. 8.2 Effectof flexural cracks on flexural bond stress In constant moment region

The bond stresses follow a distribution somewhat like that shown in Fig. 8.2(c), with the direction of the bond stress reversing between the cracks [Ref. 8.11. The net

The bending moment, and hcnce thc tensile stressf,, arc maximum at thc section at D. Evidently, if a stress f,is to bc developed in the bar at D, the bar should not be terminated at D, but has to be extended ('anchored') into the column by a certaiu

300

DESIGN FOR BOND

REINFORCED CONCRETE DESIGN

length CD. At thc discontinuous end C of the bar, the stress is zero. The difference in force between C and D is transferred to the sur~oundingconcrete through anchorage bond. The probable variation of thc n,tclro,uge bond stress u, is as shown in Fig. 8.3(b) - with a maximum value at D and zero at C. It may be noted that a similar (but not identical') situation exists in the bar segment CD of the simply supported beam in Fig. 8.l(c). An cxprcssion for an average bond s t r w u,!,, can be derived by assuming a uniform bond stmss distribution over the length L of the bar of diameter r$ [Fig. 8.3(c)], andconsidering equilibrium of forces as givcn below:

301

where zb, is the 'design bond stress', which is the permissible value of the average anchorage bond slress u,. The values specified for zbd (CI. 26.2.1.1 of the Code) are 1.2 MPa, 1.4 MPa, 1.5 MPa, 1.7 MPa and 1.9 MPa for concrete grades M 20, M 25, M 30, M 35 and M 40 and above respectively for plain bars in tension, with an increase of 60 percent for deformed bars in tension, and a further increase of 25 percent for bars in compression. The development length requirements in terms of L d / $ ratios for fully stressed bars (fa = 0.87 f, ) of various grades of steel in combination with various grades of concrete are listed in Table 5.6. It may be noted that when the area of steel A, actually provided is in excess of the area required (for& = O0.87fy)), then the actual

-

This bond stress may be viewed as the average bond stress generated over a length L in order lo develop a maximunitensile (or compressive) stressf, at a critical section; hence, this type of bond is referred to as '~cvelopmentbond'. Alternatively, this bond may be viewcd as that requircd to provide mc1,oruge for a critically stressed bar: hence, it is also refcned to as 'anchorage bond'. 8.3.1 Development Length

The term developmem length has already been introd~~ced in Section 5.9.2, in relation to restrictions on theoretical bar cut-off points. The concept of 'development length' is explained in the Code as follows:

development length required L,, may be proportionately reduced [Ref. 8.51:

id= Ld x (4L * d (4),,"id,

(8.6)

In the case of bundled bars, the Code specifies that the "development length of each bar of bundled bars shall be thatfor the individual bar, increased by 10percent for nvo bars in contact, 20 percent for three bars in contact and 3 3 percenr for four bars in contact." Such an increase in development length is warranted because of the reduction in anchorage bond caused by the reduced interface surface between the steel and the surrounding concrete. 8.4 BOND FAILURE AND BOND STRENGTH 8.4.1 B o n d Failure Mechanisms

Thc conccpt underlying 'development length' is that a certain minimum length of the bar is required on either side of a point of maxinlum rtecl stress, to prevent the bar from polling out under tension (or pushing in, under compression). However, when the requircd bar embcdmcnt cannot be conveniently provided due to practical difficulties, b e n h , hooks and mechanical anchorages can be used to supplement with an equivalent cmbcdment length [rcfer Section 8.5.31. The term rmcho~'agelength is sometimes used in lieu of 'development length' in situations where the embedment portion of the bar is not subjected to any flexural bond [Fig. 8.31. The expression given in the Code (Cl. 26.2.1) for 'development length' L,i rollows from Eq. 8.4:

-*

'

It can be seen !hat in thc case shown in Rg. 8.l(e), flenuml bond coexists with anchorage

bond, owing lo varialion of bending rnomenl'in the segment CD. 111 this book, the notation Ld is resewed for devdopnmt length of hilly stressed bars

*

(f,=0.87

fv)

For,f, < 0 . 8 7 f , , evidently

id= ~ , , ( f ~ / 0 . ~ 7 f ~ )

The mechanisms that initiate bond failure may be any one or combination of the following: break-up of adhesion bctween the bar and the concrete; longitudinal splitting of the concrete around the bar; crushing of the concrete in front of the bar ribs (in deformed bars); and shearing of the concrete keyed bctween the ribs along a cylindrical snrface surrounding the ribs (in deformed bars).

.

The most common type of bond failure mechanism is the pulling loose of the reinforcement bar, following the longitudinal splirting of the concrete along the bar embedment [Fig. 8.41. Occasionally, failure occurs with the bar pulling out of the concrete, leaving a circular hole without causing extensive splitting of the concrete. Such a failure may occur with plain smooth bars placed with large cover, and with very small diameter deformed bars (wires) having large concrete cover [Ref. 8.11. However, with deformed bars and-with the normal cover provided in ordinary beams, bond failure is usually a result of longitudinal splitting. In the case of ribbed bars, the bearing pressure between the rib and the concrete is inclined to the bar axis [Fig. 8.4(b)]. This introduces radial forces in the concrete ('wedging action'), causing circumferential tensile stresses in the concrete surrounding the bar (similar to the stresses in a pipe subjected to internal pressure)

302 REINFORCED CONCRETE DESIGN

DESIGN FOR

and tending to split the concrete along the weakest plane [Ref. 8.2 - 8.41. Splitting occurs along the thinnest sunounding concrete section, and the direction of the splitting crack ('bottom splitting' or 'side splitting') depends on the relative values of the bottom cover, side cover and bar spacing as shown in Fig. 8.4(b).

I

-.*,.,: x ">

,

:,A

-,; .$

"

.$ ! *;

i*

6 i:

BOND

303

continuous longitudinal splitting cracks [Fig. 8.51. However, in beams without stirrups, the failure due to bond can occur early and suddenly, as thc longitudinal split runs through to the end of the bar without the resistance offered by the s t i l ~ ~ ~ p s .

Flg. 8.5 Stirrups resisting tensile forces due to bond

Cb >

A simply suppol.ted beam can act as a two-hinged arch, and so calyy substantial loads, even if the bond is destroycd over the length of the bar, provided the tension bars are suitably anchored at their ends [Fig. 8.61. However, the deflections and crackwidths of such a beam may be excessive. The anchorage may be realised avcr an adequate length of embedment beyond the face of the support, andlor by bends and hooks or mechanical anchorages (welded plates, nuts and bolts, etc.).

c,

(a) bottom and side splitting cracks least of Co,C,, CSb

(b)splitting forces with deformed Flg. 8.4 Typical bond splitting crack patterns

ANCHORAGE

9I"1 .j

ANCHORAGE

Fig. 8.6 T~ed-archactlon wlth bar anchorage alone

#o

Splitting cracks usually appear on the surface as extensions of flexural or diagonal tension cracks in flexural members, beginning in regions of high local bond stress [Fig. 8.21. With increased loads, these cracks propagate gradually along the length of embedment ('longitudinal splitting') with local splitting at regions of high local bond stress and associated redistribution of bond stresses. It is found that in a uorlnal beam, local splitting can develop over 60 - 75 percent of the bar length without loss of average bond strength and without adversely affecting the load-canying capacity of the beam [Ref. 8.11. The presence of stirrups offers resistance to the propagation of

$

.* ..:

, % .

c

, ~, : , ,,, i (

. . .: ,. :, >. , ;. ,. ..

.' !

8.4.2 Bond T e s t s

Bond strength is usually ascertained by means of pull our tests or some sort of beam tests. The typical 'pull out' test is shown sche~naticallyin Fig. 8.7(a). A bar embedded in a concrete cylinder or prism is pulled until failure occurs by splitting, excessive slip or pull out. The nonlb~albond s r m g r h is computed as P/(n$L),where P is the pull at failure. $ the bar diameter and L the length of embedment. It may be noted,

DESIGN FOR BOND

304 REINFORCED CONCRETE DESIGN

however, that factors such as cracking (flexural or diagonal tension) and dowel forces, which lower the bond resistance of a flexural membcr, are not present in a concentric pull out test. Moreover, the concrete in the test spccimcn is subjected to a state of compression (and not tension), and the friction at the bearing on the concrcte offers some restraint against splitting. Hence, the bond conditions in a pull out test do not ideally represent those in a flexural member.

305

test, is bound to give a lesser (and more accurate) measure of the bond strength than the pull out strength. However, the pull out test is easier to perform, and for this reason, more co~mponlyperformed. R o m the results of such bond tests, the 'design bond stress' (pertnissible avcrage anchorage bond stress), zb,, ,is arrived at - for various grades of concrcte. Tests indicate that bond strength varies proportionately with and

a

a/@

for small diameter bars

for large dian~etcrdeformed b a s [Ref. 8.11.

8.4.3 Factors Influencing Bond Strength Bond strength is influenced by several factors, some of which have already been mentioned. In general, bond strength is enhanced when the following measures are adopted:

(a) 'pull out' test set up

shield separating bar

Another factor which influences bond strength in a beam is the depth of fresh concrete below the bar during casting. Water and air inevitably rise towards the top of the concrete niass,and tend to get trapped beneath the horizontal reinforcement, thereby weakening the bond at the underside of these bars. For this reason, codes specify a lower bond resistance for the top reinforcement in a beam.

8.5 REVIEW OF CODE REQUIREMENTS FOR BOND 8.5.1 Flexural Bond

(b) bond test on modified cantilever specimen

1

I

Fig. 0.7 Bond tests Of the several types of beam tests developed to simulate the actual bond conditions, one test set-up is shown in Fig. 8.7(b) [Ref 8.31. The bond strength measured from such a test, using the same expressiol~[ P/(n$L) I as for a pull-out

Traditionally, design for bond required the consideration of bothj7exural (local) bond stress ur and development (anchorage) bond stress u,,. However, since the 1970s, there hhs been an increased awareness of the fact that the exact value of flexural bond stress cannot be accurately comvuted (nsing like Eq. 8.3a or 8.3b) owing - exvressions . to the unpredictable and n&uiform distribution of the actual dund stress. In fact, i t is found that localised bond failures can and do occur, despite the checks providcd by Eq. 8.3. However, as explained earlier, these local failures do not impair the strength of the beam (in terms of ultimate load-carrying capacity) provided the bars are adequately anchored at their ends [Fig. 8.61. Thus, the concept underlying limit state design for flexural bond has shifted from the control of local bond stresses (whose

306

DESIGN FOR BOND

REINFORCED CONCRETE DESIGN

predicted values are unrealistic) to the development of required bar stresses through provision of adequate anchorage - at simplc supports and at bar cut-off points. The Code requirement for such a check on anchorage, in terms of development length plus end anchorage and the variation of tensile stress in the bar [Eq. 5.251, has already becn discussed at length in Section 5.9.3, and illustrated with several examples.

8.5.2 Development (Anchorage) Bond The computed stress at every section of a reinforcing bar (whether in tension or compression) must be developed on both sides of the section - by providing adequate 'development length', L,, . Such development length is usually available near the midspan locations of normal beams (where sagging moments are generally maximum) and support locations 01continuous beams (where hogging moments are generally maximum). Special checking is generally called for in the following instances only:

-

8.5.3 Bends, Hooks and Mechanical Anchorages Bends, conforming to standards are frequently resorted to in order to pmvide anchorage, contributing to the requkements of development length of bars in tension or compression. The Code (CI. 26.2.2.1) specifies that "the anchorage value of a bend shall be taken as 4 rimes the diameter. of rhe bar for each 45" bend, subject to a maximum of 16 times the diameter of the ba,". Commonly a 'standard 90' bend' (anchorage value = 8$) is adopted [Fig. 8.8(a)], including a minimum extension of 44. Any additional extension beyond the bend also qualifies to be included in development length calculations. However, for bars in compression (as in column bases), it is doubtful whether such extensions can meaningfully provide anchorage. The 90' bend itself is very effective in compression as it transfers part of the force by virtue of bearing stresses, and prevents the bar from punching through the concrete cover. When the bend is turned around 180" (anchorage value = 16$) and extended beyond by 44, it is called a standard U-fype hook [Fig. 8.8(b)]. The minimum (internal) turning radius ( r in Fig. 8.8) specified for a hook is 2 8 for plain mild steel bars and 4$ for cold-worked deformed bars [Ref, 8.51. Hooks are generally considered mandatory for plain bars in tension [refer Cl. 26.2.2.1a of the Code].

length = 8Q

307

length = 169

i.!.

.......

(a) standard 90' bend

(b) standard U-type hook

Fig. 8.8 Anchorage lengths of standard bends and hooks

In the case of stirrup (and transverse tie) reinforccment, the Code (Cl. 26.2.2.4h) specifies that cornplete anchorage shall be deemed to have been provided if any of the following specifications is satisfied: 90°bend around a bar of diameter not less than the stirrup diameter $, with an extension of at least 8 $; 135' bcnd with an extension of at least 6 $; 180' bend with an extension of at leas1 4 9. It may be noted that bends and hooks introduce bearing stresses in the concrcte that they bear against. To ensure that these bearing stresses are not excessive, the turning radius r (in Fig. 8.8) should be sufficiently large. The Code (CI. 26.2.2.5) recommends a check on thc bearing stress inside any bend, calculated as follgws:

where Fb,is the design tensile force in the bar (or group of bars), r the internal radius of the bend, and $ the bar diameter (or size o f b a r of equivalent area in case of a bundle). The calculated bearing stress should not exceed a limiting bearing stress, given by

1'5fck ,

where n is the centre-to-centre spacing between bars 1+2@/a perpendicular to the bend, or, in the case of bars adjacent to the face of the member the clear cover plus the bar diameter $. For fully stressed bars,

\

,

Accordingly, it can bc shown that the limiting radius is given by

DESIGN FOR BOND

3YO

31 1

REINFORCE0 CONCRETE DESIGN

8.7 DESIGN EXAMPLES flexural or direct tension and 244 under compressiont. When bars of two different diameters are to be spliced, the lip length should be calculated on the basis of the s~nalleidiameter. Splices in tension members shall be enclosed in spirals made of bars not less than 6 mm diameter with pitch not more than 1Omm. In the revised Code, some additional clauses have been incorporated (CI. 26.2.5.1~)to account for the reduction in bond strength with regard to rebars located near the top region [refer Section 8.4.3*]. When lapping of tension reinforcement is required at the top of a beam (usually near a co~~tinu~ussupport location or a beamcolumn junction) and the clear cover is less than twice the diameter of the lapped bar, the lapped length should be increased by a factor of 1.4. If the rebar is required to turn around a comer (as in an exterior beam-column junction), the lapped length should be increased by a factor of 2.0. This factor may he limited to 1.4 in the case of corner bars when the clear cover on top is adequate hut the side cover (to the vertical face) is less than twice the diameter of the lapped bar. When more than one bar requires splicing, care must be taken to ensure that the splicing is staggered, with a minimum (ccntre-to-centre) separation of 1.3 times the lap Icngth, as indicated in Fig. 8.9(c). It is also desirable to provide (extra) transverse ties (especially in columns), connecting the various longitudinal hars in the spliced region. In the case of bandled b a n , the lap length should be calculated considering the increased L,, [refer Section 8.3.11, and the individual splices within a bundle should be staggered.

EXAMPLE 8.1

Check the adequacy of thc anchorage provided for the longitudinal bars in the cantilever beam shown in Fig. 8.10 and suggest appropriate modifications, if required. The beam is subjected to a uniformly distributed factored load of 100 kN (total, including self-weight). Assume M 20 concrete a11d Fe 415 steel, deformed bars.

SOLUTION Preliminary check on anchorage - leneth Assuming the bars are fully stressed at the location of maximum moment (i.e., face of colutnn support), full development length LAis required for anchorage of the bars inside the column, beyond this section.

-

8.6.2 Welded Splices and Mechanical Connections Welded splices and nlechanical connections are particularly suitable for large diameter bars. This results in reduced consumption of reinforcing steel. It is desirable to subject such splices to\ tension tests in order to ensure. adequacy of strength [refer CI. 12.4 of the Code]. Welding of cold-worked bars needs special precautions owing to the possibility of a loss in strength on account of welding heat [Ref. 8.61. The Code (C1.26.2.5.2) recommends that the design strength of a welded solice should in -general be limited to 80 percent of the design strength of the bar for tension splices. Rut, weldinn of bars is yenerally adopted in welded splices. The bars to be spliced should be of the same diameter. - ~ d d & o n a ltwo or three synunetrically positioned small diameter lap bars may also be provided (especially when the bars are subjected to tension) and fillet welded to the main bars. Even in the case of 'lap splices', lap welfling (at intervals of 5g) may be resorted to in order to reduce the lap length. End-bearing splices arc permitted by the Code (CI. 26.2.5.3) for bars subject to compression. This involves square cutting the ends of the bars and welding the bar ends to suitable bearing plates that are embedded within the concrete cover.

' Columns are heqoentlv to comoression cornbincd with bendine. Hence.. it mav . . subiected . . be "

pmdn~lto cslculate the lap length, assamingflexs,al msimr, anii!mt comp~essioo. As explained in Section 8.4.3, this reduction in bond strength is strictly applicable for all situations, not only lap pin^. It may also he noted that the reduction in hand strenzth of loo reinfo~.cementma)i no; be s&fican~ in shallow members.

'

Fig. 8.10 Example 8.1

(T:)

For the tension burs (2 - 25 g a t top), L, = -2 @ = 47 g .- , [This follows fromf, = 415 MPa and T~ = 1.2 MPa x 1.6 for M 20 concrete with Fe 415 steel deformed hars; L,/g ratios are also listed in Table 5.41 = L d = 4 7 x 2 5 =1175mm. Actual anchorage length provided (including effect of the 90' bend and extension of bar beyond bend + 44 minimum extension) = 280 + (4 X 25) + 300 = 680 mm < L d = 1175 mm Not OK. For the compression barn (2 - 16 @atbottom), r, can be increased by 25 percent, whereby Ld= (47 g) x 0.8

*

=1 Ld = 47 X 16 x 0.8 = 602 nun Actual anchorage lcngth p~ovtded= 300 mm.

x

(b) Variation of steel tensile strain

(c) Strain profiles

f

'

363

The material called 'reinforced concrete' is essentially concrete, as the embedded reinforcement comprises only a very small fraction of the volume of the reinforced concrete member. As explained in Chapter 8. [Fig. 8.21, the bond stress is zero at every crack lacatiou and also midway between cracks (in the region of constant moment); elsewhere the bond stress varies nonlinearly.

(d) moment-curvature relation

Fig. 10.2 Effective flexural rigidity of a beam subject to constant moment

364

An effective cwvunrre & ,, may be defined for thc beam segment, as being representative of the mean curvature of the segment, under thc action of a constant moment M. For this purpose, the strain profile to be considered may be reasonably based on the meurr sauin profile [Fig. 10.2(c)], rather than tho strain profile at the crack location (which is obviously higher):

where E,! and E,,,, are the mean strains in the extreme compression fibre in concrete and tension steel respectively, d is the effective depth, and Elct is the effective flexural rigidity of the section. 'Flexural rigidity' E l is obtainable as the slope (secant modulus) of the monlentcurvature relationship, which in turn can be established from an average of a number of test rcsults [Ref. 10.5 - 10.11]. As shown in Fig. 10.2(d), this may be obtained variously as:

.. .. .

SERVICEABILITY LIMIT STATES: DEFLECTION AND, CRACKING 365

REINFORCED CONCRETE DESIGN

ElT EI,,

-based on the 'uncrackcd-transformed' section; -basedon thc 'gross' (uncracked) scction, ie., ignoring the presence of stccl;

EI*, I ,

-based on the 'effective' section; -based on the 'cracked-transformed' section [refer Eq. 4.151.

C1. C-2.1) is based on an earlier version of the British Code, which assumes an idealised trilinear moment-curvature relation [line OABCD in Fig. 10.31. The initial uncracked stiffness E l , and the cracked stiffness (at ultimate load) El,, are represented by the slopes of lines OA and OC respectively. Any intermediate stiffness (Elet corresponding to line OB) can be interpolated by defining'the slope of the intermediate Jine ABC in the region M,, < M < M,,. The slope of this line (which commences with the cracking moment point A) is approximated as 0.85EIc,. It follows that: M EIea = for M,, < M 5 M,, ?, + ( M - M,.)/(0.85EIc,) where p, = M , , / E I ~ , ~M,, ~ = f,18,/y, [Note that this expression for M,, is similar to Eq. 4.10, except that expression, it can be shown that

is used instead of IT]. Simplifying the above

moment A

Evidently, EIT represents the true flexural rigidity for M c M,,, and EIef represents the true flexural rigidity [or M > M,,.Whcreas El, is a constant and a property of the beam section, Elfldcpends on the load level (applied moment). It fallows that:

Thus, determining the flexural rigidity (stiffness) on thc basis of the uncracked section results in an under-estimation of the actual deflection of a reinforced concrete beam under service loads; whereas doing so on the basis or thc (fully) cracked section results in an over-estimation of the actual deflection.

Fig. 10.3 ldeallsed trilinear moment-curvature relation

I,, ' - 1.2-(Mc, /M)q

I where

10.3.4 Effective S e c o n d Moment of Area Formulation Various crnpiecal expressions for the 'effective second moment of ama' I@(for calculating short-tcrm deflections in simply supported beams) have been proposed [Ref. 10.5 - 10.1 I] and incorporated in diflerent codes. Some of these formulations are bascd on assumed transition moment-curvatorc relations [Ref. 10.5 - 10.81, whcreas the others [Rcf. 10.9- 10.111 arc based on assumed transition of strains/stresses in the region between cracks (and involve stress-strain relations and equilibrium of forces). The expression given in the Indian Code (IS 456: 2000,

q

= 1.2- Ic r / 18,

with I,, 5 I A?", The section is doubly reinforccd [Fig. 10.8(c)]. Taking moments of areas of the cracked-transformedabout the NA, I000 x (ka212+ (7 x 167.55)(kd- 35) = (8 x 457) x (126 - k 4 a 500(kd)2't 4828.85 (M)-501705.8 = 0 Solvina. kd = 27.2 nun. k = 27.2 I126 =0.2158 (b) SECTION 'AA'

(C) SECTION '8 8'

Fig. 10.8 Example 10.4

The moment coefficients prescribed by the Code (and used in Example 5.3) will be used here to determine the moments. (a) at midspan: MM,, = w,~'l12 = 5.5 x 3.4632/12 = 5.5 liNm per m width < A?, = 10.45 kNm Mnr.,+~= M , , + wLL12/10 = 5.5 + (4.0 x 3.4632/10) = 10.3 kNm per m width < A?, (b) ut end support: MI,, = ~ ~ ~ 1 ' 1=25.5 4 x 3.463'124 = 2.75 !dYm < G,, M,,D+L = (wm+ wLL)12124 = 9.5 x 3.463'124 = 4.75 kNm < M,,

-

(c) atfirst interior supporr: Mz,, = wDLl21l0= 5.5 x 3.463'110 = 6.60 kNm per m width < A?,, M,,., = MZ,, + wLL?l9= 6.60 + (4.0 x 3.463'19) = 11.93 kNm> G, Effective second monlent of area (a) at midspun: (i) due to DL: MALI,I,, Weighred average [Rg. 10.8(a)] lnfi= 0 . 7 1 ,,, ~ t o . I ~ ( I , = 0.7 x (3.4133 x 10') +0.15 (3.4133 t 3.4133) x 10' = 3.4133 x lo8 nu$ (ii) underDL+ LL: (l~fl,,,),+L= 0.7 x (3.4133 x 10') + 0.15 (3.4133 +0.7547) x l o 8 =3 . 0 1 4 5 ~10' Short-term deflection

..

A - A,, =---- 'I' [1.2M,,,- 0.2MJ (Eq. 10.3) 48EIer where M, = w1218 (i) L e toDL, M,,, = 5.5 kNm, M, = 5.5 x 3.463'18 = 8.245 kNn1

a A, = 5x(346312 ~ ( 1 . 2 ~ 5 . 5 - 0 . 2 x 8 . 2 4 5 ) ~ 1=0 ~0.73 lnnl 48~25000~(3.4133~10~) (ill due to DL t LL, M,,, = 10.3 kNm , M, = 9.5 x 3,463'18 = 14.24 kNm

(iii) due to LL alone: A, = A,, - A, = 1.58 - 0.73 = 0.85 nun

m4

SERVICEABILITY LIMIT STATES: DEFLECTION AND CRACKING 381

380 REINFORCED CONCRETE DESIGN

10.4 LONG-TERM DEFLECTION The deflection of a reinforced concrete flexural member incrcases with time, mainly due to:

. .

differential shrinkage or temperature vaiation (causing differential strains across thc cross-section, resulting in curvature); creep under sustained loading; and temperature effccts in statically indeterminate frames [Fig. 10.91.

The factors affecting shrinkage and creep arc relatcd to the environment, making of concrete and loading history; these have been described in detail in Sections 2.11 and 2.12. It may be noted that, unlike creep strains, shrinkam - and temperature strains are independent of the stress considerations in the concrctc. Furthermore, shrinkage and temperature effects aIe reversible to a large extent, unlike creep effects.

/ .

....

due to TEMPERATURE INCREASE

DECREASE

Additional factors which can contribute to increased long-term deflection (not considered here) include formation of new cracks, widening of earlier cracks, and effects of repeated load cycles.

10.4.1 Deflection Due to Differential Shrinkage In an unrestrained reinforced concrete member, drying shrinkage of concrete results in shortening of the mcmber. Howevcr, the reinforcing steel embedded in the concrctc resists this shortening to some extent, with thc result that compressive stress is developed in the steel, and tensile stress is developed in the concrete. When the reinforcement is placed symmetrically in the cross section, shrinkage does not result in any curvature of the member - except in statically indeterminate frame elements. where shrinkage results in an overall change in geometry of the entire frame, and has an effect similar to temperature [Fig. 10.91. When the reinforcement is unsymmetrically placed in the cross-section (as is usually the case in flexural members), differential strains are induced across the cross section - with the locations with less (or no) reinforcement shrinking more than the location with relatively high reinforcement. This is depicted in Fig. 10.10(a) for a simply supported beam (where the main bars are located at the bottom), and in Fig. 10.10(b) for a cantilever beam (where the main bars are on top). For convenience, it is assumed that the strain gradient is linear. The differential shrinkage causes a curvature ( q d in the member, which is in the same direction as that due to flexure under loading. Thus the shrinkage deflection enhances the deflection due to loads. C o d e Expression f o r Shrinkage Curvature

Flg. 10.9 Deflections in a statically Indeterminate frame due to temperature effectsor shrinkage Tbe combined long-term deflection due to shrinkage, creep and temperature effects may be as large as two to three times the short-term deflection due to dead and live loads. It may also be noted that, whereas excessive short-term deflections can be effectively countered by cambering, this is not generally done in the case of long-term deflections. Providing a camber to a reinforced concrete flexural member implies casting the member in such a configuration that, following the removal of the formwork, the member deflects (under dead loads) into a horizontal position. This is commonly done in the case of cantilever beams and slabs.

The shrinkage curvature rp,, (due to differential shrinkage) may be expressed in terms of the shrinkage strains E ~ J(at , the extreme concrete compression fibre) and c,, (at the level of the tension steel) [refer Fig. 10.101 as follows:

where d i s the effective depth

where The parameter, k, evidently depends, amongst other things, on the extent of asymmetry in the reinforcement provided in the cross-section. The Code (CI. C-3.1) suggests the following expression for shrinkage curvature based on empirical fits with test data [Ref. 10.21:

382 REINFORCED CONCRETE DESIGN

SERVICEABILITY LIMIT

STATES:

DEFLECTION AND CRACKING

383

sepwate the beam into sagging ('positive' curvature) and hogging ('negative' curvature) segments. This is depicted in Fig. 10.11 for some typical boundary conditions.

(a) simply supported beam

(c) fixed beam

(b) cantilevel beam

(d) propped cantilever

Fig. 10.10 Curvature due to differential shrinkage

where, e,,

-

9$1,= k4

Flg. 10.11 Relation between deflection and shrinkage curvature

Em 7

the ultimate shrinkage strain of concrete, and 0.72(y,-p,)/fi

Fig. 11.5 Variation of IS Code moment coefficients a,, o;with &!I, for simply suppotted and uniformly loaded rectangular slabs Detailing of Reinforcement The flexural reinforcements in the two directions are provided to resist the maximum bending moments M,== a, w,, 1: (in the short span) and M,,? = o; w,, 1.: (in the long span). The steel requirements at the midspan locations in strips distant from the middle strip progressively reduce with the distance from the middle strip. However, the usual design practice is to provide bars that are uniformly spacedt throughout thc span (in both directions), with a flexural resistance that is not less than the calculated maximum ultimate bending moment (M,, or M,J. Furthermore, considering any particular strip (transverse or longitudinal), the bending moment varies from a maximum value at the midspan to zero at either support [Fig. 11.41. Hence, it is possible to curtail the bars in accordance with the Code provisions explained in Section 5.9. For the special case of simply supported

OF TWO-WAY SLAB SYSTEMS

two-way slabs (torsionally unrcstl.ained), the Code (CI. D-2.1.1) suggests a sinlplified procedure for reinforcement curtailn~ent. According to this proceduie, up to 5 0 percent of the bars may be terminated within a distance of 0.11 from the support, while the remaining bars must extend fully into the supports. If the slab is truly sbrrply suppor.ted at the edgcs, there is no possibility of 'negativc' moments develo~ingmar the supports, due to partial fixity. However, it is good design practice Lo always safeguard against the possibility of partial fixity. As explained with reference to the design of one-way slabs [refer Chapter 51 this can b e achieved either by bending up alternate bars [Fig. 5.3, 5.5(b)l, or by providing separate top steel, with area equal to 0.5 times that provided at bottom at midspan, with an extension of 0.11 from the face of the support [Fig. 5.5(a)]. [The recent trend is to do away with bent up bars and instead to opt for separate layers at top and bottom. This type of detailing is illustrated in Example 11.1 [Fig. 11.141.

11.2.4 Uniformly Loaded 'Restrained' Rectangular S l a b s The Code (C1. D-I) uscs the term resrruinedslahs to refer to slabs whose comcrs am prevented from lifting and contain suitablc reinforcement to resist torsion [Ref. 11.111. All the four edges of the rectangular 'restrained' slab am assumed to bc supported (tied down) rigidly 8gninst vertical translation, and the edges may be either continuouslfixed or discontinuous. Accordingly, nine different configurations of restrained rectangular slab panels are possible (as shown in Fig. 11.6), depending on the number of discontinuous edges (zero, one, two, t h e e or four) and also depending on whether the disco~ltinuousedge is 'short' or 'long'. Panel type O corresponds to the slab with all four edges continuouslfixed, and panel type @ conesponds to the slab with all four edges simply supported [Fig. 11.61. [Incidentally, there will be several mare cases if combinations involving f i e (unsupported) edges are also consideredl.

-

-

T

continuous (or fixed) edge

-

simply Suppoltededge

1

-

I+

b+l

Fig. 11.6 Nine different types of 'restrained' rectangular slab panels 'The spacing of reinforcement should not exceed 3d or 300 nun (whichever is smaller),

427

428

REINFORCED CONCRETE

DESIGN OF

DESIGN

The torsional restraint at the corner calls for the provision of special comer reinforcement, as explained earlier [refer Fig. 11.2(c)]. The corner restraints have the beneficial effect of reducing the deflections and curvatures in the middle of the slab. Expressions for design momcnt coefficients for uniformly loaded two-way 'restrained' rectangular slabs with fixed or simply supported edge conditions, based on the classical theory of plates, are available [Ref 11.1, 11.5. 11.61. Approximate solutions based on the Rankine-Grashoff theory are also available. Modifications, to these solutions were proposed by Marcus ('Marcus correction'), whereby the momnlt

.

SYSTEMS 429

EDGE STRIP

(0.125 L)

MIDDLE STR

(0.75 1,)

-

For example, the 'Marcus correction' in the case of simply supported slabs, results in a reduction in thc dcsign moment M, by about 42 percent for 1,11, =1.0 (square slab) and 9 percent for 1J1, = 3.0, when compared to the slab with corners free to lift up [Table 11.1]; these results compare favourably with the rigorous solutions from clastic theory [Ref. 11.51. Howcver, the moment coeflicients rcconunended in the Code (CI. D-I) are based on inelasric arrri1)~vi.s(yield line analysis) [Ref. 11.12, 11.131, rathcr than elastic theory. This analysis is based on the following assumptions: the bottom stcel in either direction is uniformly distributed over the 'middle strip' which spreads over 75 percent of the span; the 'edge strip' lies on either side of the middle strip, and has a width equal to lJ8 or 1,18 [Fig. 11.71; top steel is provided in the edge strip adjoining a continuous edge (and at right angles to the edge) such that the corresponding flexural strength (ultimate 'ncgativc' moment capacity) is 413 times the comsponding illtinlate 'positive' moment capacity due to the bottom steel provided in the middlc strip in the direction under consideration; the corner reinforcement providcd is sufficient to prevcnt the formation of 'corncr Icvers', i.e., forking of diagonal yield lines near the corncrs.

.

The resulting moment cocfficicnts &+,o;* Tor 'positive' ~nomentsat midspans in the short span and long span directions respectively, and the coeficients &,, o;.for 'negative' moments at the continuous edge(s) in the two directions, for the nine different sets of boundary conditions [Fig. 11.61 are listed in Table 26 of the Code. The design factored moment is obtained as

M,, = aw,,l:

TWO-WAY SLAB

(11.9)

where w,, is the uniformly distributed factored load and 1, the effective short span and a is the applopriate moment coefficicnt.

EDGE STRIP

(0.1251),

Flg. 11.7 Basis for Code moment coeflicients for 'restrained' two-way slabs

'Positive'moment coefficients , : a

a;

The variations of the short span 'positive' moment coefficient &+,with 1,/1, is plotted for the nine types of two-way slabs in Fig. 11.8. In all cases, there is a marked increase in as IJl, increases from 1.0 to 2.0. The Code recommends a constant value of o;' for all values of 1J1,. The value of %+is obtainable from the following formula [Ref. 11.131:

a* Y = (24+2n, +1.5n~)/1000

(11.10)

where tm denotes the number of discontinuous edges. Corresponding to nd = 0, 1,2, 3 and 4, the values of o ;' are obtained as 0.0240, 0.0275, 0.0340, 0.0435 and 0.0560 respectivelyt. An expression for cht may be obtained in terms of o;' and r. lJl.r from yield line analysis [Ref. 11.12, 11.131 as follows:

-

where for a discontinuous edge

(11.12)

and the subscripts s and I denote 'short edge' and 'long edge' respectively, while the additional subscripts '1' and '2' represent the two cdges in either direction. Thus, for I n Table 26 of the Code, the specified values of a,are 0.024, 0.028, 0.035, 0.043, and 0.056 -corresponding to ,r.r = 0, 1.2, 3 and 4 respectively.

130 REINFORCED CONCRETE DESIGN

xample, for the slab panel of type '4' ("two adjacent edges discontinuous"), C,I + Csz = 2.5275. For such a case, nd = 2; hence, applying Eq. 11.10, CI, + C12 = 1 + q,+= 0.0340. Further, applying Eq. 11.1 1,

DESIGN OF TWO-WAY SLAB SYSTEMS 431 'negative' moment) at a col~tinuoussupport is 413 times the 'positive' moment capacity in the midspan region. Of course, at a discontinuous support, the 'negative' moment developed is zcroi . Accordingly, at a discontinuous support at a continuous support

(11.13)

Detailing of Flexural Reinforcement

Fig. 11.8 Variations in short span 'positive' moment coefficients with ldl, in 'restrained' two-way slabs This results in values of &+varying from 0.0356 (for r. = 1.0) to 0.0700 (for r = 2.0); the corresponding values given in Table 26 of Code are 0.035 (for r = 1.0) and : 0.069(for r = 2.0). Similuly, values of &+and c$+ can be easily obtained for any value of r I I&, in the range [1.0,2.0] and given set of boundary conditions. If thc value of lJ1, excecds 2.0, the Code (Cl. D-1.11) recommends that the slab should be treated as one-way [refer Chapter 51; the provision of the secondary reinforcernenf in the long span direction is expected to take care of the nominal bending moments that may arise in this direction.

(a)

PLAN

8

'Negative' moment coefficients a;, a, As explained earlier, the Code moment coefficie~ltshave been derived (using yield line analysis) with the basic assumption that the ultimate momnent of resistance (for

(b) SECTION 'AA'

Fig. 11.9 Detailing of flexural reinforcement in two-way 'restrained' rectangular slabs' (excluding corner reinforcement) However, as explained enrlier, the possibility of partial restraint lnlust be considered at the time of detailing.

DESIGN OF TWO-WAY SLAB SYSTEMS 433

.

The bottom steel for the design moments (per unit width) M,,.: = %+w,,l: and M ', = q+w,,12 should be uniformly distributed across the 'middle strips' in the short span and long span directions respectively. The Code (CI. D-1.4) recommends that these bars should extend to within 0.251 of a continuous edge or 0.151 of a discontinuous edge. It is recommcnded [Ref. 11.141 that alternate bars (bottom steel) should extend fully into the support, as shown in Fig. 11.9. The top steel calculated for the design moments Mrm-= n; w,,l: and Mu; = a;w,,l: at continuous supports should be uniformly distributed across the 'edge strips' in the long span and short span directions respectively. The Code (Cl. D-1.5) recommends that at least 50 percent of these bars should extend to a distance of 0.31 from the face of the continuous support, on either side. The remaining bars may be curtailed at a distance of 0.151 from the face of the continuous support, as shown in Fig. 11.9'. To safeguard against possible 'negative' moments at a disconti~mousedge due to partial fixity, the Code (CI. D-1.6) recommends that top steel with area equal to 50 percent of that of the bottom steel at mid-span (in the same direction) should be provided, extending over a length of 0.11, as shown in Fig. 11.9. In the edge strip, distribution bars parallel to that edge (conforming to the minimum requirements specified in Section 5.2) should be provided - at top and bottom - to tie up with the main bars [Fig. 11.91.

Design 'Negative' Moments at Continuous Supports In a wall-supported continuous slab system, each rectangular slab panel is analysed separately (for design moments) using the Code moment coefficients. The 'negative' moments (M,, MI) calculated for two panels sharing a common continuous edge may not be equal [Fig. 11.11] due to one or more of the following reasons:

Detailing of Torsional Reinforcement at Corners Torsional reinforcement is required at the comers of rectangular slab panels whose edges are discontinuous. This can conveniently be provided in the form of a mesh (or grid pattern) at top and bottom. 'The bars can be made U-shaped (wherever convenient) and provided in the two orthogonal directions as shown in Fig. 11.10 [Ref. 11.141. The Code (CI. D-1.8) recommends that the mesh should extend beyond the edge' over a distance not less than one-fifth of the shorter span (I,). The total area of steel to be provided in each of the four layers should be not less than:

.

0.75 A:,, if both edges meeting at the corner are discontinuous; 0.375 A:,, if one edge is continuous and the other discontinuous

Here, A:, is the area of steel required for the maximum midspan moment in the slab. It may be noted that if both edges meeting at a corner are continuous, torsional reinforcement is not called for at the corner [refer CI. D-1.101. This is indicated in Fig. 11.10. [However, this area will have some reinforcement provided anyway, because of the-'negative' moment reinforcements over supports in the middle strips and the distributor reinforcements in the edge strips.] For convenience in estimaling lengths of bars and locations of bar cut-off points, I, and I, may be taken as the spans, measured centre-to-centre of suppolis. In Fig. 11.9, the top bars and bottom brxs are shown as being separate. Alternatively, the bottom bars can be bent up to form the top steel. as shown in Fig. 5.5. 'Here, the term 'edge' refas to the face of the support.

bars at top and bottom (may be U-shaped) !

SECTION

'AA'

Fig. 11.10 Detailing of torsional reinforcement at corners the two adjacent spans are unequal: the boundary conditions in the two adjoining panels are different; the loading on one panel is different from that in the other panel. Since the Code moment coefficients are based on inelastic analysis, with a fixed ratio (413) of 'negative' to 'positive' moment capacities, no redistribution of moments

434 REINFORCED CONCRETE DESIGN

DESIGN

OF TWO-WAY

SLAB SYSTEMS 435

is pennissihle'. Hence, it is logical to take the larger factored moment (MI in Fig. 11.1 1) as the design 'negative' moment at the continuous edge.

bendlng T,

-

Hcnce, OK.

EXAMPLE 11.2 Repeat Example 11.1, assun~ingthat tile slab corners are prevented from lifting up.

Check for deflection control

SOLUTION [Refer Example 11.11: Assume D = 160 mm (which is 5 nnn less than the previous case) Assuming 8 c$ bars =, d , = 160 - 20 - 4 = 136 mm, [Iy = 136 - 8 = 128 nun 1, =4000+136=4136mm 1, a= ' 1.240 l y =5000+128=5128mm 4

f , = 0.58 x 415 x 3851392.5 = 236 MPa =, modificat~onfactor k, = 1.5 (from Table 5.2 or Fig. 3 of Code) =1 (lld),,,

-i

= 20 x 1.5 = 30

Loads on slab: (same as in Example 11.1) Factored load iv,, = 12.20 1i~lmn' Design Moments (for middle strips; 1 m width in each direction). As the slab corners are to be designed as torsionally restmined, the moment coefficients give11 in Table26 of the Code (CI. D-I) may be appliedt for I& = 1,240:

*

Short span: a, = 0.072 + (0.079 - 0.072) x 1.240-1.2 = 0.0748

+ M,,= &c

T

1.3 - 1.2

w,,1;

= 0.0748 x 12.20 x 4.136'= 15.61 kNmJm (which, incidentally, is about 15 percent less than the value of 18.36 kNm/m obtained in Example 11.1) Long span: o;= 0.056 * M a = 4 w,,l; = 0.056 x 12.20 x 4.136~= 11.69 k N d m (which is comparable to the earlier value of 11.94 k N ~ d m )

P L I N OF FLOOR

.

Design of reinforcement

F I ~11.14 . Example 11.1

R =%

- bd: -

1 5 . 6 1 ~ 1 0=~0.844 MPa 10~x136~

* As explained earlier, a check an shear is not really called for in unlfonnly loaded, wallsupported two-way slabs. This is evident from the results of this example. 'Alternatively, E q 11.10, 11.1I may beapplied.

DESIGN OF TWO-WAY SLAB SYSTEMS

441

-

10oox50.3 = 150.7 334 Maximum spacing permitted = 3 X 136 = 408 mm, but < 300 mm. 20 = 41-4.589~0.714/20] = 0.206 x 3 100 2x415, a (A,,), ,e,d = (0.206 x 10") x 1000 x 128 = 264 mm2hn 1000~50.3 3 Required spacing of 8 $ bars = = 191 264 Maximum spacing permitted = 3 x 128 = 384 mm, but c 300 mm 8 4 @ 150 cjc (short span) Provide 8 4 @ 190 c/c (long span) The detailing is shownin Fig. 11.15. =sRcquired spacing of 8 $bars =

L

-

-

i

I

PLAN

Check for deflection control p, .= 0.2465 r f, = 0.58 x 415 x 334/335 = 240 MPa 3 modification factor k, = 1.55 (from Table 5.2 or Fig. 3 of Code) (lid,.,= 20 x 1.55 = 31 4136 (l/d)PPPv!dad = - = 30.4 < 3 1 -Hence, OK. 136 Corner Reinforcement As the slab is designed as 'torsionally restrained' at the corners, corner reinforcement has to be provided [vide CI. D-1.8 of the Code] over a distance 1.J5 = 830 mm in both directions in meshes at top and bottom (four layers), each layer comprising 0.75 A,,,.. 150 3 spacing of 8 $bars = - = 200 c/c 0.75 Provide 8 -$ @ 200 clc both ways at top and bottom at each corner over an area 830 mm x 830 mm, i.e., 5 bars U-shaped in two directions, as shown in Rg. 11.15. ~~

L 160

T SECTION '88'

Fig. 11.15 Example 11.2

M , ,- 11.69x106 = 0.714 MPa RY=bd: - 1 0 ~ x 1 2 8 ~

~

EXAMPLE 11.3

The floor slab system of a two-storeyed building is shown in Fig. 11.16. The slab system is supported on load-hearing masonry walls, 230mm thick, as shown. Assuming a floor finish load of 1.0 kN/m2 and a live load of 4.0 khVm2, design and detail the multipanel slab system. Use M 20 concrete and Fe 415 steel. Assume mild exposure conditions.

442

REINFORCED CONCRETE

DESIGN

DESIGN OF

SOLUTION

a Effective depths

Thc slab system [Fig. 11.61 has two axes of symmeiry passing tlu.ough the centre, owing to which the number of different slab panels to be designed is four:

one short edge discontinuous

I

@

TWO-WAY SLAB SYSTEMS 443

i

d, =US-20-4=111mm d, = I l l - 8 =103mm

~IJI,=510314111 = 1.241 Panels .%and S3: 3.0 m x 5.0 m clear spans a 1, 3000 + 100 = 3100 mtn a d, = 31001(23 x 1.5) = 90 mm [The continuity effect is only partial in the case of panel SJ. Hence, it is appropriate to consider a basic lld ratio which is an average of the simply supported and continuous cases, i.e., (20 + 26)/2 = 23.1 Assuming a clear cover of 20 nun and 8 @bars,D = 90 + 20 + 812 = 114 mm ProvideD = 115 mm d, =115-20-4=9lmm a effective depths d, =91-8 =83mm

--

=3000+91=3091mm =5000+83 = 5083mm

a effective spans

Loading on slabs self-weight of slab

for S t , S 4 =2.875kN/m2 f o r S 2 , S ,

@ 2 5 k ~ / r nxo0.135rn= ~ 3.375kN/m2

@ 2 5 x 0.115

finishes @ 1.0 k?J/mZ live loads @ 4.0 !&Vrn2 =, Factored load w,, = 1.5x(3.375+1.0+4.0) = 1 2 . 5 6 d ~ / m ~f 0 r S t , S 2

lSx(2.875+l.0+4.0) =1!.8f'/t~/m'

for S3,S,

Design Moments (using Code moment coefficients for 'restrained' slabs) Referring to Table 26 of the Code, or alternatively applying Eq. 11.10 - 11.12, the q ' (for 'positive' moments in the middle following moment coefficients strip) in the different panels are obtained as: (0.037, 0.028 panel St

4,

I

Flg. 11.16 Floor slabsystem - Example 11.3 Slab thicknesses: based on deflection control criteria Panels SIand S4: 4.0 m x 5.0 m clear spans 1. = 4000 + 150 = 4150 tm

Assuming a clear cover of 20 mm and 8 g bars, D..l07+20+8/2=131mm Provide D = 135 mm

'The width of the continuous suppo1.1(230 mm)is less !ha" 1/12 of the clear span (4000112 = 333 mm); hence, the effective span is to be take11 as (clear span + d)or (centre-to-celttre distance between supports), whichever is less [refer CI. 22.2 of Code].

DESIGN OF TWO-WAY SLAB The coefficients for the 'negative' moments in thc vwious continuous edge strips are easily obtained as a- = 4/3at The co~~esponding design (factored) monlents M,, = a w,,l? in the various pallets are accordingly obtained as follows:

.

SYSTEMS

445

long sprm: M:, = 3.16 k N d m =, (As),cYd= 108.4 mm2/m =, reqd spacing of 8 $ bars = 464 ium - to be limited to 3Oy = 249 nun

M,i. = 5.27 k N d m =) (A,,),eqd = 184.4 mn12/m =, reqd spacing of 8 $bars = 272 nun - to be

limited to 3r1, = 249 m m

Design 'negative' mon~entsa t common supports The 'negative' moments at the continuous edges, as obtained from the Code coefficients, arc uncqual - as shown in Fig. 11.17(a). In all such cases, thc design 'negative' moment is taken as the larger of the two values obtained from either sides of the support. The design moments so obtained are show11 in Fig. 11.17. Flexural reinforcemel~trequirements (a) 'negative' moments (kNm/rn)at continuous edges for each panel wherehk = 20 MPa& = 415 MPa, b = 1000 mm Panel SI: d* = 111 tnm, d, = 103 mm shortspan: M : ~ = 7.85 k N d m (A&,,=

203.6 mm2hn

a reqd spacing of 8 $ bars = 247 nun M,; = 10.47 k N d m a (A,,),+ = 275.5 imn2/m a reqd spacing of 8 $bars = 183 nun long span: = 5.94 k N d m a (A,,),, = 165.2 mm2/m

MA

reqd spacing of 8 $ bars = 304 mmlimited to 300 mm M,; = 7.92 k N d m a (A,,),,, = 223 mm2/m

to be

=, reqd spacing of 8 $ bars = 225 mrn Panel S2:d, = 91 mm, d, = 83 mm * short span: MA = 6.32 !&dm a (A,,),cd = 201.6 mm2/m

reqd spacing of 8 $bars = 249 m n Mt; = 9.62 k N d m a (A,,),d = 315.5 1 m ~ 1 m a reqd spacing of 8 $ bars = 159 mm

(b) final design moments (kNm/m)

Fig. 11 . I 7 Design moments

- Example 11.3

DESIGN OF Panel 5'3: rl., = 91 mm, 4 = 83 mm d?o,r sparr: M:? = 6.77 kNmlm 3 (A&d

a ieqd spacing of 8 $bars = 232 mm = 346 mm2/~n

a reqd spacing of 8 $bars = 145 mm long span: MlY = 3.95 Wmlm a (A&d = 136.5 mm2/m

M, =5.27 k N l m 3

SYSTEMS 447

The area of steel required at the other corners, where torsional reinforcement is required, is half the above requirement however, provide 3 nos 8 nun $at top nnd . . bottom. -~ ~ ~The size of the mesh is 620 mm X 620 lNn at the junctioll of Sz and S3,where onc edge of the corner is discontinuous. TIlc size of [he mcsh is 0.2 x 41 11 = 820 nlm at the junction of S,and S,. Provide 3 nos 8 mm 4 bars at top and bottom The detailing is shown in Fig. 11.18. [Note: Tile slab panels satisfy the limiting I/d iatios for deflection control; this may be verified.]

-

= 216.8 mm2/m

M; = 10.47 Wm/m a (A&d

=)

TWO-WAY SLAB

rcqd spacing of 8 4bars = 368 nun - to be limited to 3d, = 249 nun = 184.4 molz/m

a reqd . spacinz . - of 8 b bars = 272 nun - to he linited to 3dy = 249 nun ~

~

Panel S4: d, = 111 mm, d, = 103 mrn short spfln: M& = 7.22 kNm11na (A,,),,,,l = 186.7 mm2hn

k

5230

li;-

3-8Battopand bottom (4 layers. each corner) at >K-2615

3 reqd spacing of 8 $ bars = 269 nun M,: = 9.62 kNmlm 3 (A,,),&= 251.9 mm%n 3 reqd spacing of 8 $bars = 199 m n ~ long span: M; = 5.09 Wmlm a (A&d = 140.9 mm2/m

reqd spacing of 8 4 bars = 357 mm - to be limited to 35or 300 mm M,;. = 7.92 WmJm a (A,,) ,.e, = 223 mm21m

,

a reqd spacing of 8 $bars = 225 mm Detailing of Reinforcement Based on the requirements of reinforcement calculated above, the detailiug of flexural reinforcen~entin the various middle strips and edgc strips is shown in Fig. 11.18. For practical convenience, only two different bar spacings (220 nun and 150 mnl) are adopted (except for slab S3 for M,Yx, for which a spacing of 145 is used). The detailing is in conformity with the requirements specified in C1. D-1 of the Code, and satisfies the requirements of minimum spacing. * Nominal top steel (50 percent of bottom steel) is provided at the discontinuous edges -against possible 'negative' moments due to partial fixity. Torsiortal irirrforcented at conrers As required by the Code, the reinforcement is provided in the form of a mesh, extending over a distance of 0.21.. beyond the face of the supporting wall. The bars are provided as U-shaped (i.e., with the mesh extending over top and bottom). At the extreme corner of the slab system, required spacing of 8 mm 4 bars = 413 x 220 = 293 mm -over a distance of 0.2 x 3103 = 620 mm. =) Provide 3 nos 8 mm $ U-shaped bars in both directions at the extreme corner of the slab system -over a distance 620 mm x 620 mm.

PL A N

(showing main bars) Fig. 11.%a)l

[for lengths of bars, refer SP : 34 or

SECTION

,

IAA'

Fig. 11.18 Detalllng of mult~panel slab system - Example 11 3

.--

DESIGN Panel SJ: d, = 91 mm, d, = 83 1nm short span: ML. = 6.77 kiimnlm =, (A,,),,,

= 21 6.8 mm2/m a k q d spacing of 8 $bars = 232 nun MI; = 10.47 k N d m a (A,,) ,, = 346 mn12/nl * reqd spacing of 8 $ bars = 145 mm

,

long span: M$ = 3.95 k N d m

M,; =5.27 ! d W m

* (A,,),,,= 136.5 mm2/m * reqd spacing of 8 $ bars = 368 m m - to be

* (A,,),ad= 184.4 =1reqd

limited to 3 4 = 249 mm mm2/m

spacing of 8 $bars = 272 mm - to be limited to 3d, = 249 mm

Panel Sd: d, = I l l lnm, d, = 103 mm shorr span: M:x = 7.22 kNmJm (A,,,),,,, = 186.7 mm2/ni' reqd spacing of 8 $bars = 269 mm M,;x = 9.62 N m l m a (A,),,, = 251.9 nnn2/m

OF TWO-WAY

SLAB

SYSTEMS

The area of steel required at the other comers, where torsional reinforcement is requimd, is half the above requirement - however, plovidc 3 nos 8 mm $ at top and bottom. Thc size of the mesh is 620 lmn x 620 mtn at the junction of S, and S,, where one edge o l the comer is discontinuous. The size of the mesh is 0.2 X 4111 = 820 mm at the junctioli of S,and S1. Provide 3 nos 8 mm bars at top and bottom. The detailing is shown in Fig. 11.18. [Note: The slab pancls satisfy the limiting l/dratios for deflection control; this may be vcrified.]

p

3-8qattapand bottom (4 layers, at each corner)

5230*F2615-x

I

*

reqd spacing of 8 g bars = 199 mm long span: M& = 5.09 kiimfm =, (A&,, = 140.9 mm2/m

a reqd spacing of 8 $bars = 357 mm - to be limited to 3dyor 300 mm . .

M;, = 7.92 liNmlm a (A,,),,,, = 223 mm2/m reqd spacing of 8 $bars = 225 mm

Detailing of Reinforcement Based on the requirements of reinforcement calculated above, the detailing of flexural reinforcement in the various middle strips and edge strips is shown in Fig. 11.18. For practical convenience, only two different bar spacings (220 mm and 150 mm) are adopted (except for slab S3 for M,;*, for which a spacing of 145 qm is used). The detailing is in conformity with the requirements specified in C1. D-l of the Code, and satisfies the requirements of minimum spacing. e Nominal top steel (50 percent of bottom steel) is provided at the discontinuous edges -against possible 'negative' moments due to partial fixity. Torsional reinforcernertt a t corners As required by the Code, the reinforcement is provided in the form of a mesh, extending over a distance of 0.21, beyond the face of the supporting wall. The bars are provided as U-shaped (i.e., with the mesh extending over top and bottom). e At the extreme corner of the slab system, required spacing of 8 mm $ bars = 413 x 220 = 293 mm -over a distance of 0.2 x 3103 = 620 mm. =1 Provide 3 nos 8 mm $U-shaped bars in both directions at the extreme corner of the slab system -over a distance 620 mm x 620 mm.

447

(showing main bars) bars, refer SP : 34 or Fig. 11.9(a)]

PLAN

[for lengths of

Fig. 11.18 Detailing of multl panel slab system - Example 11.3

DESIGN OF TWO-WAY SLAB SYSTEMS 449 11.2.6 Design of Circular, Triangular a n d Other S l a b s

where the llotations are exactly as mentioned earlier [Fig. 11.191

Design r?rontenrs (assuming

Nan-rectangular slabs, with shapes such as ci~culnr,triangular and trapezoidal, are sometimes encountered in structural design pactice. Rectangular slabs, supported on three edges or two adjacent edges, am also mct mil11 i n practicc. Classical solutions, based on the elastic theory, are available in thc case of rectangular and circular plates, oniformly loaded [Ref. 11.11. Nonrectangular slabs are sometimes designed by considering the largest circle that can be inscribetl within thc boundaries of thc slabs, and treating these slabs as eqllivalent circular slabs [Ref. 11.16]. Morc accurate analyses of stresses in nonrectangular and olher slabs are obtainable from the computer-based finite difference method [Ref. 11.61 and finite element method [Ref. 11.71. Yield line analyses provide simple and useful solutions for slabs of all possible shapes and boundary conditiol~s[Ref. 11.8 - 11.10]. Some of the standard solutions for a few typical cases are given here (without derivation). In all these cases, slabs are assumed to bc subjected to nr~ifornll~ distributed loads w (per unit area).

V

= 0):

+ = wnz/16 ('positive' a t centre) M,,nax M,,, = (-) wn2/8.; M& = 0 (at edges)

M,I.,,,

[TO,,

is

(11.17~) (11.17d)

=

near the supports, in the radial direction]. o~lhmonal

circumferential bars

:

Circular Slabs, Simply S u p p o r t e d [Fig. I I . 191 Elastic theory

centre) W

Moment in radial direction M, = -[(3 I6 Moment in circunlIerentia1 direction where

---

+

(11.15a)

- ,-2)1

Fig. 11.19 Circular slabs, simply supported and isotropically reinforced

M, = x [ 0 2 ( 3 + v ) - r 2 (I t3")] I6 (11.15b)

radius of the circular slab; radios where moment is deternlincd (0 5 r < a); V Poisson's ratio - may be taken as zcro in thc case of reinforced concrete. Maximum moments (at centre): M,,,,,, = = lwa"l6 (11.15~) The two-way reinforcemcnt may bc provided by mcans of an orthogonal mesh wit11 isotropic reinforcement [Fig. 1 l.l9(b)l: Providing radial plus circumferential rcinforcernent [Fig. 1 I.l9(c)l is also (theoretically) a solution; however, this is not convenient in practice, as the radial bars necd lo bc specinlly welded at the centre. o r.

Yield line theory (assmning isotropic reinforcement) Collapse load w,, = 6M,,,/nZ =? desigtl momcnt M,, = w,,n2/6 = w,,02/5.333) (which is less than the elastic theory solotion: M,,

.

Yield llne theory (assuming isotropic reinforcement) z Collapse load w,, = 6( M,,;+ M i )/a tile elastic moment at the support is twice that at the centre, it is desi~ableto provide MI,; : Mi,; in the ratio 1 : 2 (11 18a) Accordingly, for design, M,: = (+) waZ/18 M,;= (-)waZ/9 (11.lRb) Equilateral Triangular Slabs, Simply S u p p o r t e d [Fig. 11.20(a)]

(11.16)

Circular Slabs, Fixed a t the Edges Elastic theory

M,= -1v[ n Z ( l + v ) - r 2 ( 3 + v ) ] 16

(1 1.17~1)

(8)

(b)

Fig. 11.20 Equilateral triangular slabs, isotropically reinforced

DESIGN OF TWO-WAY SLAB SYSTEMS

Yield line theory (assuming isotropic ~einforcemne~~t)

* Collapse load w,, = 72 M,,,llZ wherc I is the length of one side =) design moment (at midspan) M,, = w,,12172

Rectangular Slabs, Three Edges Fixed and One Edge Free [Fig. 11.21 (b)]

-

Yleld line theory Let P M.R,Y/M,,x,x ix = ~ " i , x / ~ , , ; , x

Equilatera! Triangular Slabs, Two Edges Simply Supported and One Edge Free [Fig. 11.20(b)] Yield line theory (assuming isotrop~creinforcement) Collapse load w,,= 24 M,,,l12 where 11sthe length of one s ~ d e design moment (at midspan) M, = ~ ~ 1 ~ 1 2 4 (11.20)

i ~ - M,,R.~/M,&

-

-

..

=I

w,,%(3-a,)

.

Rectangular Slabs, Three Edges Simply Supported and One Edge Free [Fin. 11.21ia)l ~. . Yield llne theorv Let P M t r ~ , ~/M,,R,x and R ldlx ; the X- and Y- directions are as indicated in Fig. 11.21(a). De.sign moments:

451

Design moments:

where ..

M,J;

-

a, 4(-

7

*

6(iy la:)

(whichever is greater)

(1 1.22)

6(l+iy)

2)/K1

.

'.

By suitably selecting M, ix and iy,all the design n~omentscan be determined. This is illustrated in Example 11.5.

'Rectangular Slabs, Two Adjacent Edges Fixed and the Other Two Free [Fig. 11.221 By suitably selecting p (which can even be taken as unity), design moments M , , and M,,y = p MfrXcan be deterl~ned.

k G y x d ivM+w

Fig. 11.22 Rectangular slab, two edges fixed and the other two free

Fig. 11.21 Rectangular slabs, supported on three edges

452

REINFORCED CONCRETE

DESIGN

DESIGN OF TWO-WAY

SLAB

SYSTEMS 453

Yield line theory e

Using the same notations as in the previous casc [mler Fig. 11.221, design moment: MiIMX=

.

where

(whichever is greater)

(m

a3= - I)/K~ By suitably selecting p , ix and i,, all the design moments can be dete~mined

EXAMPLE 11.4 Design a circular slab of 3.5 m diamcter to cover an onderground sump. The slab is simply supported at the periphery by a wall 200 mm thick. Assume a finish load of 1.0 kNlrn2 and live loads of 4.0 !di/m2. Use M 20 coucretc and Fe 415 sleel. Assumc ntildcnposure conditions. SOLUTION o

Clear span = 3500 - (200 x 2) = 3100 mm Assuming a slab thickness of 100 mm,with 20 111111 clear covcr (mild exposure condition) and 8 mm $ bars (in an orthogonal mesh), average effective depth d = 100 - 20 - 8 = 72 mm effective span (diameter) = 3100 + 72 = 3 172 mm effective radius n = 317212 = 1586 m m Loads: (i) sell weight @ 25 k ~ l xd0.10 m = 2.5 ~ l d =1.0 " .(ii). finishes (iii) live loads = 4.0 " w = 7.5 w / m 2 3 Factored load w,,= 7.5 x 1.5 = 11.25 kN/m2

=

31b0

200

I

'I

200

PLAN

Pig. 11.23 Circular slab - Example 11.4 EXAMPLE 11.5 Determine the design moments in a square slab (4 m x 4 m), with three edges continuous and one edge free, subject to 3 uniformly distributed factored load XI,,=10.0 kN/mnz. Assume the slab to be isotropically reinforced. Also assume that the 'negative' monlent capacity at the continuous support to be equal to that at midspan in either direction.

.

SOLUTION Applying the yield line theory solution [Eq. 11.221 with Ix = 1, = 4.0 m, R = 1, p = i x = i y = 1,

Design ,,$ontents (assuming yield line theory with isotopic reinforcement) M,, = 1v,,a2/6 = 11.25 x 1.586~16=4.72kNm1m

-" =x[l-41-(4.598x0.910/20)]

= 0.267 x 10" 2x415 = (0.267 x 10.') x lo3 x 72 = 192 mm21m 3 A,, =, required spacing of 8 mm c$ bar = 50.3 x 10'1192 = 262 mm Maximum spacing allowed = 3d = 3 x 72 = 216 nun Provide 8 mm $I 210 clc both ways at bottom, as shown in Fig. 11.23

w.&3-a,)

100

e

3

M ; ~=

6(iy + a 1 ) w,,$ a: ----6(l+ i,)

= 0.0338 w,,l; = 0.0353 w,,l:

(greater.)

DESIGN

11.3.3 S l a b s S u p p o r t e d o n Flexible B e a m s

- C o d e Limitations

As an alternative to the idealised assumption as continuous slabs supported on walls, the Code (CI. 24.3) suggests that slabs monolithically colmected with beams may be analysed ns members of a continuous fiuw~cwork ivitlt the s~rpports, taking into nccowt fhe stiffnness of srrch sripports. This suggestion becomes significant in situations wherc the supporting beams are not adcquatcly stiff. It may be noted that the ACI Code had made such a treatment mandatory for all (flexible) beam-supported two-way slabs, as far back as in 1971, and had altogether dispensed with the use of moment coefficients for s r ~ hslabs. Anothei significant change introduced in thc 1971 version of the ACI Codc was the unification of the design methods for all slabs supported on columns - with and without beams, including flat slabs. This is considered to be an advancement as it ensures that all types of slabs have approximately the same reliability (or risk of failuret ) [Ref. 11.61. Some other codes, such as the Canadian codc [Rcf 11.181, have also incorporated these changes, but retain the moment coefticicnt method as an alternative for slabs supported on walls or sriflbeaws. In the IS Code, such a procedore, based on the concept of 'equivalent frame', is prescribed for f i t slobs. This method [CI. 31 of the Code] follows closely the extensive research undcrtnkcn i n this area, since 1956, at thc University of Illinois, USA [Ref. 11.111. Howcver, unlike the ACI and Canadian codes, the IS Code is yet to extend the 'equivalent frame' concept of analysis to beam-supported slabs. The problem of designing slabs on flexible beams has thcrcfore not yet been satisfactorily addressccl by the IS Code. This information is also not generally available in standard Indian books on reinforced concrete design. In the sections to follow, p~ocetlumsfor analysis and design of beam-supported slabs are described - in line with the by-now-well-eslablished ACI concept of unified procedures for all slabs, supported on columns, with or without beams.

11.3.4 T h e 'Equivalent Frame' C o n c e p t As mentioned in Section 11.1.3, in the case of beam-supl~ortedtwo-way slabs, 100 percent of the gravity loads on the slabs are tmnsmitted to the supporting columns, in both longitudinal and trallsverse directions (see Fig. 113(b)). The mechanism of load transfer from slab to colunms is achieved by flcxurc, shear and torsion in the vwious elements. The slab-beam-column system behaves integrally as a three-dimensional system, with the involvefnent of all the floors of the building, to resist not only gravity loads, but also lateral loads. However, a rigorous threc-dimensional analysis of the

-

"

is separated from the design of beams (and columns), as in the case of wall-supported slabs. Thc remaining pan of the structore, comprising a tluee-dimensional skeletal

'

It is reported thnr the application of the 'mornem coeficicnt' procedure to bean-sopported slabs results in inorc conservative designs, with the resuk rliar such slabs turn out to be significantly stronger than beanlless (flat) slabs, given the same gravity lpnds and material gmdes [Ref. 11.6].

OF TWO-WAY SLAB

SYSTEMS 457

framework of beams and columns, is separated for convenience, into (twodimensional) plane frames in the longitudinal and transverse directions of the building. As the integrally cast slab also contributes to the strength and stiffness of the beams, the beam members are considered as flanged beams (T-beams, L-beams), with portions of the slab acting as the flanges of these beams; this concept was explained in Chapter 9. However, when the beams arc flexible or absent, it is not appropriate to separate the slab design from the beam design. In using the concept of a plane frame comprising colunlns and slab-beam members at various floor levels, fundamentally, the slab-beam member should consist of the enfirefloor member (slab and beam, if any) tributary to a line of columns forming the frame. This is illustrated in Fig. 11.24(a) and (b), which show how a building structure may be considered as a series of 'equivalent (plane) frames', each consisting of a row of columns and the portion of thefloor system tributafy to it. The part of the floor bound by the panel centrelines, on either side of the columns, forms the slabbeam member in this plane frame. Such 'equivalent frames' must be considered in both lonaitudinal and transverse directions, to ensure that load transfer takes place in both directions [Fig. 11.24(a)l. The eauivalent frames can now be analysed under both gravity loads and lateral loads usiig the procedures mentioned in Chapter 9. The primary differmce between the frame in Fig. 9.l(b) and the one in Fig. 11.24(b) lies in the width of the slab-beam member and the nature of its connections with the columns. Whereas in the conventional skeletal frame, the full beam is integral with the column, and the rotational restraint offered by the columl at the joint is for the entire beam (with both beam and column undergoing the same rotation at the joint), in the 'equivalent frame', the column connection is only over part of the slab-beam member width, and hence the flexural restraint offered by the column to the slab-beam member is only partial. Thus, the rotation of the slab-beam member along a transverse section at the column support will vary, and will be equal to the colunm rotation only in the inunediate vicinity of the column. This, in turn, results in torsion in the portion of thc slab transverse to the span and passing through thc column (i.e., a cross-beam running over the colonul). In the elastic analysis of the plane framc in Fig. 9.l(b), il was shown (in Section 9.3) that several approximations can be made, subject to certain limitations. Similar approximations can also be made in the present casc. For example, for the purpose of gravity load analysis, it is possible to sinlplify the analysis by applying the concept of substitute ,frames. Accordiagly, instead of analysing the full 'equivalent frame' [Fig. 11.24(b)], it suffices to analyse separate partial frames [Fig. 11.24(c)], comprising each floor (or roof), along with thc columns located immediately above and below. The columns are assunled to be fixed at their far ends [refer CI. 24.3.1 of the Codc]. Such substitute frame analysis is permissible provided the frame geomctry (and loading) is relatively synunetrical, so that no significant sway occurs in the actual frame.

458

REINFORCED CONCRETE

DESIGN OF TWO-WAY SLAB SYSTEMS 459

DESIGN

internal equivalent frame (Y direction)

/......

.... 12

.~ ......, ~

(a) Floor Plan -definition of equivalent frame

),),!' ,/

~ ~

haif middle strip

(b) typical internal equivalent frame (X - direction)

(c) substitute internal equivalent frame (X - direction)

Fig. 11.25 Moment variations in a two-way slab panel Fig. 11.24 The 'equivalent frame' concept

A

co~urnnstrip

-

460 REINFORCED C O N C RE T E DESIGN

Variations of M o m e n t s in a Two-Wav Slab P a n e l Although thc horizotital rnembcr in thc 'eqoivalcnt partial frame' in Fig. 11.24(c) is modclled as a very wide beam (i.e., slab with or without beam along the cohmm line), it is actually supported on a very limited width. ~ e h c e the , outer portions of thd metqber arc less stiff than the part along thc column line, and the distribution of moment across thc width of the meniber is not uniforrii - unlike the beam in the convcntional planc framc [Fig. 9.l(b)]. The probablc variations of moments in a typical panel of the 'equivalent frame' are shown in Fig. 11.25. The variation of bending moment in the floor meniber along the span, under gravity loads is sketched in Fig. 11.25(b). Such a variation - with 'negative moments' ncar thc supports and 'positive momcnts' in the neiglibourhood of the midspan - is typical in any beam snbject to iiniformly distributed loads. In Fig. 11.25(b), M,, denotes thc total 'negative' moment in the slab-beam member along the support line AB (cxtending over the lull width of the panel), and M.l denotes thc total 'positive moment' along thc tniddlc linc EF of the panel. These nloments are distributed acmss the width of the panel nononiformly, as sketched in Fig. 11.25(c). The actual variation along AB or EF ((markcd by the solid line in Fig. 11.25(c) depcnds on scveral factors, such as the span ratio 12/1,, relative stirfncss o l bcam (if any) along colunm lines, torsional stiffness of lransversc beams (if any), etc. The actual moment variation is very difficult to predict exactly, and hence suitable appmximations need to be made. This is generally achieved by dividing the slab panel into a column swip (along the column line) and two half~ n i d d esrrips [Fig. 11.25(a)1, and by suitably apportioning the total moment (Mabor Me/) to these strips with the assumption that the moment within each strip is uniform. This is indicated by the brokei~lines in Fig. 11.25(c), and is also clearly shown in Fie. 11.2S(d). ~~, When bcatns arc provided along the column linc, the bcam portion is relatively stiffer than the slab and resists i major sharc of lhc inomcnt at the section. 111 this case, the moment has to be apportioned betwcen thc beam part and the slab part of the slab-beam rnembcr as indicated in Fig. 11.25(e). The calculations involved in the design procerlure are given in the ticxt section, which follows thc unifier1 procedure of desigl~for all types of column-supported slabs - with or without beams (i.e., including flat slabs).

-

11.4 DESIGN O F COLUMN-SUPPORTED SLABS (WITH I WITHOUT BEAMS) UNDER GRAVITY LOADS 11.4.1 C o d e P r o c e d u r e s B a s e d o n t h e Equivalent F r a m e C o n c e p t

Two-way slabs supported on colilmns includc jinr /,tares [Fig. 1.121, flnr slabs IRg. 1.131, ivaflc (ribbed) slabs [Fig. 1.1 11, and solid slabs with beams along the colormi lines [Fig. 110(b)]. Such slabs may bc designcd by any procedure which satisfies the basic cvnditions of equilibrium and geometrical compatibility, and the Code reqoiremcnts of strength and serviceability. Specific design procedures have

acco~dingto the Code (CI. 31.1) as follows:

I

Thc above definition is very broad and encompasses the various possible colummsnpported two-way slabs mentioned earlier. Flat slabs may have an edge beam, whicl; helps in stiffening the discontinuous edge, increasing the shear capacity at the critical exterior column supports and in supporting exterior walls, cladding etc. Furthermore, they have a favourable effect on the minimum thickness requirement for the slab (see Section 11.4.2). As mentioned earlier, the Code procedure is based on the elastic analysis of 'equivalent frames' [Fig. 11.241 mder gravity loads,.and follows closely the 1977 version of the ACI Code LRef 11.111. However, unlike the unified ACI Code procedure, there is no elaboration in the IS Code (CI. 31) for the particular casc of two-way slabs with b e a m along column lines as in Fig. 1.10(b). Thc design procedures described hereinafter will not only cover the provisions in the prevailing IS Code, but also include provisions in other international corles [Ref 11.18, 11.19] to cover the case of two-way rectangular slabs supporied on flexible beams. These Code procedures are an outcome of detailed analyses of results of extensive tests, comparison with theoretical results based on the theory of plates, and design practices employed successfully in the past. The interested reader may also refer to the background material for the Code procedures presented in Refs. 11.20- 11.22. The following two methods are recommended by the Code (Cl. 31.3) for determining the bending moments in the slab panel: either method is acceptable (provided the relevant conditions are satisfied): f

1. Direct Design Method 2. Equivalent Frame Method These methods are applicable only to two-way rectangular slabs (not one-way slabs), atid, in the case of the Direct Design Method, the recommendations apply to the gravity loading condition alone (and not to the lateral loading condition). Both methods are based on the 'equivaleqt frame concept' (described in Section 11.3.4). The slab panel is defined (CI. 31.1.lc'of the Code) as that part orthe slab bounded on cach of its four sides bv the colunm centrelincs. Each slab vanel is divided into colrrrnn strips ant1 middle srri/~.r[Rg. 11.261. A 'colunm strip' is defined (CI. 31.1.1a of tlie Code) as a design strip having a width equal to the lesser oi0.251, or 0.251, on cach side of the colunln centrcline, and includes within this width m y dmp panel or beam (along the column linc). Hcrc, I , atid 12' me thc two spans of the

' '

Othcr design procedures bascd an yield line analysis and finite elernern nnalys~sarc alrl acceptable, provided they satidy all the requirements mentioned cal.lier. In general, subscript 1 identities parameters in the direction where twmcurs a e being determined, nad subsctipt 2 relates to the pelpendiculai direction.

462

REINFORCED CONCRETE

DESIGN

rectangular panel, measured centre-to-centre of the column supports. The 'middle strip' is defined (CI. 31.1.lb) as a design strip bound on each of its sides by the column strip.

DESIGN OF TWO-WAY SLAB SYSTEMS

463

Both methods require the values of sevcral rclative stiffness parametcrs in order t o obtain the longitudinal and transverse distribution of factored moments in thc design strips. For this purposc, as ~vellas for determining the dead loads on thc slab, it is necessary to assunlc, initially, the gross section dimensions of the floor system (and the columns). Thcsc dimensions may need to be modified subsequently, and the analysis and design may thercforc necd to be suitably revised.

beam widlh not less than column width

u

column

Fig. 11.26 Column strip and middle strip in a slab panel While considering an 'equivalent frame' along a colurm~linc, the slab width i2 consists of two half nuddle strips flanking one c o l u ~ l ustrip, ~ as shown in Fig. I1.2S(a) and Fig. 11.26. When monolithic beams are provided along the colunn~lines, the effective (flanged) beam sections (which form part of the column strip) are considered to include a portion of the slab on either side of the beam, extending by a distance equal to the projection of the beam above or below the slab (whichever is greater), but not exceeding four times the slab thickness, as shown in Fig. 11.27 [Ref. 11.18]. In cases where the beam stem is very short, the T-beam may be assumed to have a width equal to that of the column support [Fig. 11.27(c)]. The Direct Design Method (described in Section1 1.5) and the Equivalent Frame Mcthod (described in Scction 11.6) for gravity load analysis differ essentially in the manncs of deternuning the distribution o i bending moments along the span in the slab-bcam member [Fig. 11.25(b)]. The fanner uses moment coefficients (similar. in concept to the simplified Code procedure for continuous b e a m and one-way slabs see Sections 5.6.1 and 9 . 3 , whcrcas the latter requires an elastic partial frame analysis. The procedure for apportioning the factored moments betwee~lthe nuddle strip and the colurntl strip (or between the slab and the beam when beams are present along h e colu~miline) is identical for both desigu methods.

(c)

(d)

Fig. 11.27 Definition of beam section

11.4.2 Proportioning of S l a b T h i c k n e s s , Drop Panel a n d Column Head

Slab T h i c k n e s s The thickness of thc slab is generally governed by deflection control criteria. [Shear is also an important design criterion - especially in f i r plurcs (slabs without beams and drop pancls) and at extcrior column supports]. The calculation of detlcctions of two-way slab systen~sis quitc complex, and recoursc is often made lo empirical rules which limit maximom spanhlepth ratios as indircct measures 01deflection control. For this pnrposc, the Codc (CI. 31.2.1) recon~mendsthe same /Id ratios prescribed (in CI. 23.2, also refer Section 5.3.2) for flexural members in general, with the following important differences:

I

464 REINFORCED

CONCRETE

-

DESIGN

'I

the longer span should be consideredt (unlike the case of slabs supported on walls or stiff beams, where the shorter span is considered); for the purpose of calculating the modification faclor k, [Table 5.21 for tension reinforcement, an overage percentage of steel across the whole width of panel should be considered [Ref. 1 I.11]; When drop panels conforming to CI. 31.2.2 arc !!or provided around the column supports, in flat slabs tlle calculated lld ratios should be further rcduced by a factor of 0.9: the minimum thickness of the flat slab should bc 125 mm

Slab Thickness Recommended by other Codes Other empirical equations for maximum spanklcpth ratios have been established, based on the results of extensive tests on floor slabs, and have bee11supported by past experience with such construction under normal values of oniform loading [Ref. 11.18, 11.191. Thus Ref. 11.8 recommends equations 11.25 to 11.26~1for the mininlum overall thichiess of slabs necessary for the control of dcflcctions. If thesc minimum thickness requirements are satisfied, deflections need not be computed. Thcse equations are also applicable for two-way slabs supported on stiff beams. Howcvcr, thcsc thicknesses may no1 be tile most cconomical in all cases, and may cvcn be inadequate for slabs with large livc to dead load mbos. In the calculation of spanldcpth mtio, the clear span 1, in the longer. dimtion and the overall depth (thickness) D arc to be considered. For flat plates and slabs with column capitals, the minimum overall thickness of slab is:

D 2 (1. (0.6 + f,I 1000)] 130 (11.25) However, discontinuous edges shall be provided with an edge beam with stiffness ratio, &, of not less than 0.8, failing which the thickness given by Eq. 11.25 shall be increased by 10 percent. For slabs with (Imp panels, the minimum tbickricss of slab is: D 2 [I,, (0.6 +& I lOOO)] 1 [30(1+(2.rdIl,)(D,,-D)ID}] (11.26) where x * / (1,,12) is the smaller of the values determined in tlle two directions, and xdis not greater than lJ4, and (Dd-D) is not larger than D. Far slabs with beams between all strpports, the minimom tl~icknessof slab Is: (I 1.26a) D 2 [I, (0.6 +f, / 1000)l I (30 + 4pab,,,) where G,,,Is not greater than 2.0. This limit is to ensure that with heavy beams all around the panel, the slab thichiess does not become loo thin. In the above equations, D, overall thickness of drop panel, nun; xd dimension from face of column to edge of drop panel, mm; f , e characteristic yield strength of steel (in MPa):

-

'

These IS Code provisions apply to 'Flat Slabs' ss defined in CI. 31.1. However, it is not clear from the Code whether they apply to slabs with flexible beam between all supports as such slabs are not specifically covered by the Code. R e t 11.8 does cover such slnbs also.

DESIGN OF TWO-WAY

SLAB SYSTEMS 465

p (clear long spany(c1ear short span); Q,,, 2 average value of & for all beams on edges of slab panel; ab= 'beam stiffness parameter', defined as the ratio of the flexural stiffness of the beam section to that of a width of slab bounded laterally by the centmline of the adjacent panel (if any) on each side of the beam. Refering to Fig. 11.27, E2b 11, a,,= c*ab=~ 1, " EC1,

(11.27)

where Ib is the second moment of area with respect to the centroidal axis of the gross flanged section of the beam [shaded area in Fig. 11.27(a), (b), (c) and (d)] and I, = 1 ~ ~ 1 is 1 2the second moment of area of the slab. The minimum thickness for flat slabs obtained from Eq. 11.25 are given in Table 11.2 Table 11.2 Minimum thicknesses for two-way slabs without beams belween interior column supports (Eq. 11 2 5 )

*

Thickrtess to be

** Edge beam must satisfy a 2 0.80.

.

.

Drop Panels Thc 'drop panel' is formed by local thickening of the slab in the neighhourhood of thc supporting column. Drop panels (or simply, drops) are provided mainly for the purpose of reducing shear stresses around the column supports. They also help in reducing the steel requirement for 'negative' moments at the column supports. [Also refer Section 11.7 for calculation of reinforcement at drop panels]. The Code (CI. 31.2.2) recommends that drops should be rectangular in plan, and have a length in each direction not less than one-third' of the panel length in that direction. For exterior panels, the length, measured perpendicular to the discontinuous edge from the column centreline should be taken as one-half of the corresponding width of drop for the interior panel [Fig. 11.28(a)l. The Code does not specify a minimum thickness requirement for the drop panel. It is, however, recommended [Ref. 11.18, 11.1?] that the projection below the slab should not be.less than one-fourth the slab thickness, and preferably not less than 100 mm [Fig. 11.28(b)l.

'

This may be interpreted as one-sixth of the centre-to-centre dimension to colurrls on either side of the centre of the column under consideration, as depicted in Fig. 11.28(a).

DESIGN OF TWO-WAY C o l u m n Capital The 'column capital' (or columrr iread), provided at the top of a colmm, is intended primarily to increase the capacity of the slab to resist punching shear [see Section 11.8.21. The flaring of the colunm at top is generally done such that the plan geometry at thc column head is similar to that of the column.

SLAB

SYSTEMS 467

The Codc (CI. 31.2.3) restricts the structurally useful portion of thc coiun~ncapital to that portion which lics within the largcst (inverted) pyranud or right circular cone whjch bas a vertex angle of 90 degrees, and can be included cntirely within the outlines of the column and tlic column head [Fig. 11.28(b)]. This is based on the assun~ptionof a 45 dcgree failure planc, outside of which enlargements of the support are considered ineffective in transferring shear to thc column [Ref. 11.1 I]. In the Dil-cct Design Method, the calculation of bending lnornents is based o n a clear span I,,, measured facc-to-face of the supports (including column capitals, if any) bur nor less t l ~ a r0.65 ~ tirues rltc panel spar. in the directioil under consideration [Cl. 31.4.2.2. of the Code]. When the colulnn (support) width in the direction of span exceeds 0.35 I,, (lo be more precise, 0.1751, on either side of the column centreline), the critical section for calculating the factored 'negative' moment should be taken at a distance not greatcr thao 0.17511from the centre of thc colun~n(C1.31.5.3).

F

11.4.3 TRANS ER OF SHEAR AND MOMENTS TO COLUMNS IN BEAMLESS TWO-WAY SLABS .......&drop panel !

column caoital (head)

I

(a) PUN

only that portion of column capital lying within a pyramidlcone with veriex angle of 90' to be considered in design calculations

clear span

1

",f,,b

slab-,

M"!" /

=Y

4

(I 1.2Xa)

where

4ui)

column

J-

Shear forces and bending ~nomentshave to be transferred bctween the floor systcrn and the supporting columns. In slabs without beams along column lines, this needs special consideratior~s. The desigu moments in the slabs are computed by franle a~lalysisin the case of the Equivnletlt Frame Method, and by empirical equations in the case of the Direct Design Method.. At any colu~nnsupport, the total unbalanced moment must bc resisted bv the colunms above :ind below in .urouortion lo their relative stiffnesses . [Fig. 11.29(a)]. In slabs without beams along the colunu~line, the transfer of thc unbalanced moment from the slab to thc column takes place partly through dircct flexural stress? and partly through development of non-uniform shear strcsses nmund the colun,n head. A part (M,,b) of thc unbalanced moment M,, can be considered to the tra~lsfenrd by flcxure and the balancc (M,,,.) through eccentricity of shear forces, as shown in Fig. 11.29(b) and (c). Thc Codc rccommcndation (Cl. 31.3.3) for the apportioning of M,,b and M,,, is bascd on a study described in Ref. 11.23:

column capital clearspan2

effective capital size

(b) SECTION 'AA'

Fig. 11.28 Drop panel and column capital

Hcre, cl and c2 arc the rlimcnsions of thc equivalent i-ectmgulal colorno, capital or bracket, measured in thc direction moments ace bcing dcter~nincd and in thc transverse direction, respectively, and d is the effcctivc depth af the slab at the critical section for shear [refcr Scclion 11.8.2]. For squarc and round colunlns, c , = c2 , aud y = 0.6.

DESIGN OF column M2 > M,

TWO-WAY SLAB

SYSTEMS 469

The width of the slab considered effective in resisting the moment M,,b is taken as the width bctween lines a distance 1.5 times slabldrop thickness o n either side of the column or column capital [Fig. 11.29(b)l, and hence this strip should have adequate reinforcemcnt to resist this moment. The detailing of reinforcement foy moment transfer, particultuly at the exterior column where the unbalanced moment is usually the largest, is critical for the safety as well as the performance of flat slabs without edge beams. The critical section considered for moment transfer by eccentricity of shear is at a distance dl2 from the periphery of the column or column capital [Fig. 11.29(c)l. The shear stresses introduced because of the moment transfer, (assumed to vary linearly about the centroid of the critical section), should be added to the shear stresses due to the vertical support reaction [refer Section 11.8.21. 11.5 DIRECT DESIGN METHOD

11.5.1 Limitations The Direct Design Method (DDM) is a simplified procedure of determining the 'negative' and 'positive' design moments (under gravity loads) at critical sections in the slab (slab-bcam member), using empirical moment coefficients. In order to ensure that these design moments are not significantly different from those obtained by an elastic analysis, the Code (CI. 31.4.1) specifies that the following conditions must be satisfied by the two-way slab systems for the application of DDM. 1. There must be at least three continuous spans in each direction. 2. Each panel must be rectangular, with the long to short span ratio not . exceeding 2.0. 3. The columns must not he offset bv more than 10 percent of the span (in the direction of offset) from either axis between centrelines of successive columns'. 4. The successive spa1 lengths (centre-to-centre of supports), in each direction, must not differ by more than one-third of the longer span. 5 . The factored live load must not exceed three times the factored dcad load (otherwise, moments produced by pattern loading would be more severe than those calculated by DDM).

(b) moment transferred

by flexure

MU"

There is an additional limitation prescribed by other codes [Ref. 11.18. 11.191, with regard to the application of this method to slab panels supported on flexible beams on all sides. The following condition, relating the relative stiffnesscs of the bcams in the two .perpendicular directions, needs to be satisfied in such cases: .

= (1 - 73M"

0.2 (C)

moment transferred through shear

<

abll: < 5.0

(1 1.30)

ad? -

whcrc abis the beam stifiesspammetcr. (defined by Eq. 11.27), and thc subscripts 1 and 2 rcfer to the direction moments are being determined, and transverse lo it,

Fig. 11.29 Transfer of unbalanced moment from slab to column

'

If the column offsets result in variation in spans in the transverse direction, the adjacent transverse spans should be averaged while carrying out the analysis [CI. 31.4.2.41.

DESIGN OF TW O - W A Y SLAB SYSTEMS

470

471

REINFORCED CONCRETE DESIGN

q - l+(l/a,)

whele respectively. Ref. 11.18 also limits the live Ioadldead load ratio to < 2.0. Furthermore, the loads are assumed lo be gravity loads, uniformly distributed over the entire panel. Total Design Moment f o r a S p a n In any given span, I,, the roral (factor-ed)design moment M, for the span, is expressed as [refer CI. 31.4.2.2 of the Code]:

11.5.2

where w,, 3 factored load per unit area of the slab; I,, 3 clear span in the direction of M,,measured face-to-face of columns' , capitals, brackets or walls, bur nor less than 0.651,'; 1, i length of span in the direction of M,;and 12 3 length of span transverse to II. With reference lo Fig. 11.25(b), it can be seen that considering statics, the absolute sum of the 'positive' and average 'negative' design momenta in the slab pawl must not be less than M, The expression for M, [Eq. 11.31] is obtained as the tnaxin~um midspan static moment in an equivalent simply supported span I,,, suhjccted to a m~iforlnlydistributctl total load W = w,,(121,,),where I&, is the erfective panel area on which w,,acts. When drop panels are used, the confributio~lof the additional dead loa,d (due to local thickening at drops) should be suitably accounted for; this is illustrated in Example 11.7. Longitudinal Distribution of Total Design Moment The typical variation of moments (under gravity loads) in the slab, along the span, has already been introduced in Fig. 11.25(b). The Code i-ecommendation (CI. 31.4.3.3) for the distribution of the calculated 'total design moment', M,, bctween critical 'negative' momcnt sections (at the face of equivalent rectangular supports) and 'positive' moment sections (at or near inidspan) is as depicted in Fig. 11.30. Inrevior spun: * 'negative' design moment Ma- = 0.65M0 (11.32a)

ac s

C K,

-

E

(1 1 14)

(X)/K,

(1 1 15)

sum of fl~xuralstiffi~cssesof columns meeting at the exterior joint: and

KSb flexural s t i f f k x oI the slab (or slab-beam member) in the direction nloments arc calculated (is., along span I!), at the exterior joint.

r; 'positive' design moment Mai = 0.35M, Exferior spun: * 'negative' design moment at exterior suppo~tM&, = (O.hSlq)M,

*

(1 1.32b) (1 1.33~1)

'negative' dcsign nlonient at info.io,- support Ma,,,, = (0.75 - O.lOlq)M, (1 133b) 'positive' desigrl rnorncnt Mot = (0.63 - 028/q)M,

(1 1 . 3 3 ~ )

Circda 1 rl~nre~ta~~gltlnr colutnn S L I ~ ~ O arc I ~ Sto be rrcated as equwalent squinelrectangulw supports having the same area (CI. 31 4 2 . 3 of the Code). This condition is imposed in order to pt.event undue rrduction in the design moment when the columns w e long and namw in cross-sectioa or have large brackets or capitals

'

[Ref. I I . I I 1 .

!,,

,p

;i!

4

Fig. 11.30 Distribution of M, into 'negative' and 'positive' design moments (longltudinal)

11.5.3

a

.,, ',,

The Code (CI. 31.4.3.4) permits a limited readjustment in the apportioned design moments - but by no more than 10 percent, because of the approximations and limitations inherent in the Direct Design Method - provided the total design moment M, for the panel is not less than the value given by Eq. 11.31. No additional redistribution of moments is permitted. For the purpose of calculating the flexural stiffness ( K d of the slab-beam member and that of the column (Kc), it is permitted [Ref. 11.11] to assume that the members are prismatic (i.e., having uniform cross-section throughout their lengths). This is underlying done purely for convenience, and is in keeping with the s~l~plifications DDM. This assunlption implies that the contributions of drop panels, colunln capitals and brackets may be neglccted. Furthennore, the increase in the second moment of area of the slab-beam member, bctween the column centwline and column face, may be neglected. With these simplifications, the flexural stiffness K (of the column or slab-beam mcmber) is simply obtained as: K = 4E,I/l

(11.36)

where E,

DESIGN OF

-

short-term modulus of elasticity of concrele (for the grade applicable to the element); I = second moment of area (considering thc gross section r ); and 1 E appmpriatc centre-to-centre span.

Thc Code (under CI. 31) covers only flat slabs, and as such Eq. 11.32 to 11.35 are not specifically meant for use in the case of two-way slabs with flexiblc beams between column supports. However, other codes [Ref. 11.181 do permit thc usc of DDM for the latter case mentioned above. For two-way slabs with beams, for interior spans, Eq. 11.32(a), (b) arc applicable. For extcrior spans, the distribution of M, for slabs with beams may be made according to the factors given in Table 11.3 -case (2).

TWO.WAY SLAB

SYSTEMS 473

apportioned transversely to the design strips of the panel (column strip (cs) and half middle strips (hms) in flat slabs, and the beam part and the slab part when there are beams) at all critical sections. This procedure of transverse distribution of moments is common for both Direct Design Method and Equivalent Frame Method..

IS Code Recommendations The Code recommendations for flat slabs (CI. 31.5.5) on studies reported in Ref. 11.24: 'Negative' moment a t exterior support:

h this regard are based mainly

Table 11.3 Moment factors for end span [Ref. 11.18]

*

halfmiddle strip: Mhis,erl

.

if col. width < 0.751,

I"

= 0.5(1- b,/l,)~&,

otherwise

(11.37b)

where bc, is the width of the column strip [Fig. 11.251 'Negative' nzoment a t interior support: * col~rnmstrip: M& = 0.75 M&,

* r

11.5.4

Apportioning of Moments to Middle Strips, Column Strips and Beams

As mentioned earlier [Fig. 11.25(c), (d)], the calculated design 'positive' and 'negative' slab moments in the panel in the longitudinal direction have to be

'

In the case of the slab-bearn member, the width of the section should be taken as the full panel width, i2.

= 0.125 Mojo,

'Positive' momentfor all spans: * column strip: M : = 0.60 M :

* It may be noted that when the slab is stiffened with beams along the column Lines, the calculation of thc second moment of area I must logically include the contlibution of the portion of the beam projecting below (or above) the slab. The calculations related to flexural stiffness arc required for the putpose of determining the parameter a, [Eq. 11.351, required for the moment factors in the end span [Eq. 11.331. An alternative and simpler scheme for end spans (with moment coefficients independent of q), suggested in Ref. 11.18, is shownin Table 11.3. [As already indicated above, this Table also covers the case of end spans of two-way slabs with beams between all columns.]

halfnriddle strip: M,&,

halfnriddle strip: MIA,,, = 0.20 M :

The transverse distribution of moments for a typical exterior slab panel is depicted in Fig. l1.31. The moments indicated are assumed to be uniformly distributed across the width of the respective design strips. In the case of a panel with a discontinuous edge in the direction of M, (span 10, such as in the external equivalent frame shown in Fig. 11.24(a), the design of the halfcolumn strip adjoining and parallel to the discontinuous edge, as well as the middle strip in the panel, depends on whether a marginal beam (with depth > 1 S D 3 or wall is supporting the slab at the edge. If such a stiffening of the edge exists, the bending moments in the half-column strip should be taken as one-quarter of that for the first interior column strip, and the moments in the middle strip as twice that assigned to the half-middle strip corresponding to the first row of interior columns (CI. 31.3.2b and 31.5.5.4~of the Code).

DESIGN OF TWO-WAY SLAB SYSTEMS 475

474 REINFORCED CONCRETE DESIGN

eXter10r d c o ~ u m n

NW

bL

I V

coiumn strip

interior column b

o.iZSM~i~,, -

AW

Funhennore, at interior calunu~s,at least one-third of the reinforcement for the total negative moment shall be located in a band with a width extending a distance of 1.5D from the sides of the column. Similarly, reinforcement for lhc total negative moment at exterior column is plt~cedwithin such a bandwidth. This is to facilitate the ~ransfcrof uobnlanced tnoment to the column by flexure. (ii) Middle strip ,rtorrto,rs: At all critical sections. the ponion of tile ncgativc and positive ~nomcntsnot resisted by the column stl.ip is assigned proportionately to the two half middle snips on either side of thc column st+ A full middle strip in a pancl has moments assigned to its two halves fiom the equivalent fr.uncs on either side (Fig. 11.32). The middle strip adjacent to and parallcl wit11 an edge supported by a wall must be desigt~edfor twice tile moment assigned to its interior half ponion fonning p.art of the equivalellt hame along the first raw of interior supports (Fig. 11.32).

Fig. 11.31 Transverse distribution of bending nlometlts in a typical exerior panel

Canadian Code Recommendations A !?taresimplified scheme for transverse distribution of moment, which accounts for the ability of slabs to redistribute monents, is given in the Canadian code [Ref. 11.18]. Slabs are highly statically indeterminate and usually gwatly under-reinforced. This leads to the forination of yield lines along sections of peak moment, effecting considerable moment redistribution to sections of lesser lnolncnt (see Section 9.7 and Rg. 9.12). This inherent ability of the slab gives the designer considel.abIe leeway in adjusting the moment field and designing the reinforcement accordingly, subject to static equilibrium conditions and requirelknts of arengtll and serviceability being met. Reflecting this flexibility, for slabs without beams, the Canadian code gives a rnngc of values for the column strip share of moment, from which the designer can choose an appropriate value: the balance is apportioned to the middle strip. For slabs with beams, the distribution is between the beam part and the slab part, the proportions being dependent on the beam stiffitess ratio and the span ratio 12/1,. This procedure is applicable to both D i m t Desigr Method and Equivalent Frame Method. These provisions are summarised below: (o) re gala^ slabs tvill~orcrbenms This includes flat plates and slabs with drop panels andlor column capitals, which may or ,nay not have edge beams along the discontinuous edges. (i) Cobmtr~mil? ntomems: The calamn strips are designed to resist the total negative or positive moments at the critical sections (given in Fig. 11.30 for DDM) multiplied by an nppropriate factol. within the following ranges: * Negative moment at interior colunu~0.6 to 1.00 Negative moment nt exterior columu 1.00 Positive moment in all spans 0.50 to 0.70

Flg. 11.32 Factored moments in middle Strips (b) Regrrlnr slobs willr benats betwcert oll srtppor.1~ In this case, the slab-beam member is divided into the beam palt (see Figs.11.25e and 11.27) and the slab part whieit is the portion of the member outside the beam part.

,-, - - -

~

~~

The beam shall be designed to resist the following fractions of the total negative and nositive nmnents at tile critical sections (given in Fig. 11.30 foi. DDM): Negative momcot at interior columns and positive moment in all spans (11.40) a ", , ~ r l +.. i l ,../ . l,~~l Negative mnonlenr at an exterior colunul - 100 PeSCellt

.

~

.

DESIGN OF

Here abl is not taken larger than 1.0. The beam must also resist moments due to

loeds directly applied on it and not considered in slab design such as weight of walls and the beam rib. The negative moment i~cinfomementat exterior suppon must be [)laced within a band of width extending a distmc 1.50 past the sides of the column or the side of the beam web. whichever is larger. (ii) A40,rwm irr slabs: The po~tionof the negative and positive moments not resisted by the beam is assigned to the slab parts outsidc the bcam. The slab reinforcement for the negative lllomcnt at interior supports is onifondy distributed over the width of the slab. Positive moment reinforcements may also bc distributed uniformly. 11.5.5 L o a d s on the E d g e B e a m

TWO-WAY SLAB SYSTEMS

477

done. M ,, gets restricted to 2 x 0.67 T,, 1 2 / (1, - cd, and the 'positive' moment in the span has to be correspondingly adjusted to maintain the value ofM, [Eq. 11.311. The design of the edge beam for torsion should conform to the requirements described in Chapter 7. Some torsion can also be expected to occur in the transverse beams at the interior columns, due to unbalanccd moments in the pancls on the two sides of these beams. However, such torsion is generally negligible (except in exceptional cases), and hence not considered in design. o,

,

torsional member (stmess K,)

The IS Code (C1. 31.3.2) describes a procedure for the design of 'marginal beams' (edge beams) having an overall depth greater than 1.5 times the slab thickness or walls supporting a flat slab. According to this procedurc, the slab portion in the 'halfcolumn strip' adjacent to the edge beam (or wall) should be designed to resist onequarter of the design motnent assigned to the 'first interior column strip'. The edge beam or wall should he dcsigncd to c m y the loads acting directly on it (if any) plus a unifonnly distributed load equal to one-quarler of the total load on the slab panel [also refer Section 11.5.81. 11.5.6 Torsion in E d g e Beam

Although the IS Code does not offer any specific rcco&endation for torsion in the transverse beatn at the exterior edge, it is evident that some of the 'negative' moment at the exterior edge of the panel (M,,) will be transferred to the column by torsion in the edge beam [Fig. 11.331, and the balance will be transferred to the column through flexure at the colomn-slab connection. The edge beam, therefom, has to be designed as a sporrrlr-el beam subjected to a torsional moment distributed along its length (in addition to the bending moments and shear force due to the Loads indicated in Section 11.5.5). For this purpose, it may be aswnicd, conservatively, that the entire 'negiltive' design nioment at the exterior support, M,,, is unifortnly distributed over thc width of the design strip, h, as shown in Fig. 11.33. This results in a linear variation of twisting moment in the edge beatn, with a zero value at the midspan (panel centreline) and a maximum value at the face of the column, as illustrated. If the maximum torque (T,",) exceeds the cracking torque (T,,)of the edge (spandrel) beam, torsional cracking will occur; there will be a reduction in the torsional stiffness and a consequent redistribution of moments' [Ref. 11.291 resulting in a relaxation in the induced torque. For this reason, if T,,,, exceeds 0.G7Tc,, Ref. 11.18 permits the use of a maximum factored torque of 0.67T,, provided a corresponding readjustnient is made to the 'positive' moment in the span. If this is

Fig. 11.33 Torsion in edge beam Torsional member and s t i f f n e s s The transverse torsional member (at the edge as in Fig. 11.33 or over an interior coluinn) is assunled to have a cross section consisting of the larger of: (i) a portion of the slab having a width equal to that of the column, bracket, or capital measured in the direction 1, plus that part of the transverse beam (if any) above and below the slab; and (ii) the beam section as defined in Fig. 11.27. For the calculation of the torsional stiffness of this member, needed for the Equivalent Frame Method, the torsional property C of the section (refer Section 7.2.3) is needed. The computation of an exact value of C for a flanged section being very difficult, it suffices to obtain an approximate value for C by subdividing the section into rectangles such that the summation of the C values of the component rectangles results in a maximum value [Fig. 11.341. This is done by subdividing in such a way as to minimise the length of the common boundaries. The expression for C [refer Section 7.2.31 is accordingly obtained as:

where x and y are the short and long dimensions of the rectangular p a t [Fig. 11,341 'refer Section 9.7.3 ('torsional plastic hinge').

478

REINFORCED CONCRETE DESIGN

kx,+

Adopt larger Ccompuled for cases 1 and 2 Fig. 17.34 Computation of torsional property Cfor a flanged edge beam

11.5.7 M o m e n t s in C o l u m n s a n d Pattern Loading As explained in Section 114 . 3 , colu~ru~s (and walls) built ~nonolithicallywith the slab n~ustbe designed to resist the unbalanced moments transferred from the slab. The total lnolnent transferred to the exte~iorcolumn is the same as the 'negative' design monlcnt ( M A , ) at thc exterior support, computed by the factors in Fig. 11.30 and Table 11.3 for DDM and obtained by frame analysis in EFM. The negative momelm at faces of supports computed by these factors for the DDM correspond to the action of full factored live load plus dead load, whereas the maximum unbalanced moment at the interior support would occur under pattern loading. Hence, in the case of the interior c o l u ~ m ,the Code (CI. 31.4.5.2) recommends the use of the following empirical expression' for the total unbalanced (factored) moment to be resisted by the column:

The total unbalanced moment wansferred to the column (exterior or interior) should be distributed to the columns above and below the floor under consideration. io direct proportion to tl~cirrelative stiffhesses. Furthermore, in the casc of beamless slabs, as explained in Scction II.4.3, a fractioo, M,,, = yM,,,of thc u~~balanced moment is transferred by flexure of a width o f slab, and the balance, M,,,, = M,, - M,,a,is transferred by shear. The design of the slab should account for the resultiug tlcxural and shcar stresses.

%,.,

3y:

.I:i

Effects of Pattern L o a d i n g on Slab M o m e n t s The moments at &tical sections in the slab, computed in DDM using the factors i n Fig. 11.30 also correspond to the aodication of full factored load (dead load plus live load) on all spans. Due to 'pauern loading' ( i t . , occunence of dead load on all spans ~ ~ - ,. it - is ~ossible and live load only on certain critical snans). that tl~c'~ositive'bcndinnmoments could excced the calculated values for full loading (on all spans), in a n extreme case, by as much as 100 percent [Ref. 11.11]. However, t h e ~ eis no likelihood of a significant increase in the calculated 'negative' monlent at the support. because the loading pattern for maxi~numinomell1 for such a case rcquires full factored loads to bc coosidcrcd on both spans adjoining the support. If the relative stiffness of the columns, m e a s u ~ by ~ I the pararnetcr a, (defined by Eq. 11.35), is high, such excess of the maximum 'l~ositive' moment under pattern loading (over that calculated with full loading on all spans) is low. The Code (CI. 31.4.6) prescribes that if a, is nor less than a specified value &, ,,,,,, then the possible illcrease in the dcsign 'positive' moment may be ignored. IT, however, the colurmis lack the dcsircd minimum relative stiffness, i.e., cl, < cr,, ,,,,,,, then the calculated 'positive' design lmomcnts should be incrcased by a factor, S, defioed as follows:

..

ac z (X Kc I X KJ

where

w,,,~,,,w,,,,, I design (factored) dead and live loads per unit area on the longer

span; w , , ~ , -design dead load per unit area on the shorter span; 1,.

1; E lengths transverse to the direction of M,,in the long span

and short span respectively; I,, , J; e lengths of clear spans (measured face to face of supports) in the direction of M,,, in the long span and short spa11 respectively; a , -relative column stiffness parameter; and Kcand K, are flexural stiffnesses of column and slab respectively.

'This exoression for M,,has been derived for the case of two adjacent unequal spans, with full

where w,, and w,, denote rcspcctively the characteristic (unfactorcd) dcad load and live load p e onit ~ area. The value of a , ,,,,,,depends on the W ~ L I Lload V L ~ratio, the lill, span ratio, as well as the beam stilfness parameter ab(defined by Eq. 11.27). The values of a,, ,,,#,, listed in Table 17 of the Code arc for Flat Slabs (i.e, without beams) for which ab= 0; and these are same as the values given as thc first set io column 3 of Table 11.4. More comprehensive valucs for G,,,,,,, covering values of ahother than zero, as given in the 1984 revision of the Canadinn Code are included in Table 11.4. For intermediate values of w,,LJIY~,.,12/1, nnd nh,linear interpolation may be resol-tcd to. The morc recent (1994) revision of the Canadian Code [Rcf. 11.181 restricts the use of DDM to cascs with wLL/~vOL< 2 and dispenses with the factor 6,.

.. ,,

DESIGN OF TWO-WAY SLAB

Table 11.4 Values of a,,,,,,,

SYSTEMS 481

above two extreme conditions of 'adequate stiffness' ( a , , 12/11 Z 1.0) and zero stiffness ( a , , 12/1, = O), as recommended in Ref. 11.18. Accordingly,

This means that, in such cases, the beams framing into the column transmit only part of the shear from panels to the column, and the balance shear is transmitted by the slab directly to the columns in two-way shear. In such cases, the total shear strength of the slab-beam-column connection has to be checked to ensure that resistance to the full shear occurring on a panel is provided. This involves the checking of the shear strength of the slab-beam part around the column perimeter as in the case of flat slabs (Section 11.8.2). It may be noted that, in addition to the shear due to slab loads, beams must also resist shears due to factored loads applied directly on the beams. The application of the Direct Design Method is illustrated in Example 11.6. 11.6 EQUIVALENT FRAME METHOD

11.5.8 Beam Shears in Two-way Slab System with Flexible Beams Shear in two-way slabs without flexible beams along colunln lines is discussed in Section 11.8. When there are beams, in addition to designing the slab to resist the shear force in it, the beam must also be designed to resist the s h c a it is subjected to. As mentioned earlier, the design of two-way slabs supported onflexible beams is not adequately covered in the Code, and hence there arc no specific reconmendations for determining the design shear forces in such beams. In general, the Code (CI. 24.5) suggests that thc desigu shear in b e a m supporting solid s l d ~ sspannir~gin two directions nt right orfglcs mrrl supparring lrnifornlly distributed loads may be computed as that caused by loads in tributary areas bounded by 45' lines drawn from the corners of the panels, as explained in Chapter9 [Fig. 9.51. However, such an idealisation is meaningful only if the supporting beams can be considered to be "adequately stiff', which, as explained earlier, is indicated by the condition a,, 12/1, 2 1.0. In the other extreme, when there are no beams, a,, 1,/l1 = 0, the 'beam' carries no load and the full sllcar in the panel is transmitted by the slab to the column in two-way action [rcrer Section 11.81. For the case of flexible beams, with 0 < a,, 12/1, < 1, the beams fiatning into the colutmts transmit only part of the shcar, and the balance of thc sllcar in thc panel is assumed to be transmitted by the slab to the column. A simplc means of evaluating the shear component in :he beam in such a case is by applying linear interpolation between the

The 'equivalent frame method' (EFM) of design (also called Elastic Frame Method) of two-way beam-supported slabs, flat slabs, flat plates and waffle slabs is a more general (and more rigorous) method than DDM, and is not subject to the limitations of DDM [refer Section 11.5.1]. Furthermore, under lateral loads, recourse has to be taken to design by EFM. The 'equivalent frame' concept has already been introduced in Section 11.3.4. Such a concept simplifies the analysis of a three-dimensional reinforced concrete building by subdividing it into a series of two-dimensional (plane) frames ('equivalent frames') centred on column lines in longitudinal as well as transverse directions [Fig. 11.241. The 'equivalent frame method' differs from DDM in the determination of the total 'negative' and 'positive' design momcnts in the slab panels - for the condition of gravity loading. However, the apportioning of the moments to 'column strips' and 'middle strips' (or to beam and slab) across a panel [refer Section 11.5.41 is common to both methods. 11.6.1 Equivalent Frame for Analysis The bending moments and shear forces in an 'equivalent frame' are obtained in EFM by .In c.l:~\licn~l;dysis'. Such an annlysis should gcncrally bc pe!lbrmr.d on the :ntirc nl;lne fr;lnic 1C1 3 1 5 l I.a.) d l h : (.hJcI. Iloucvcr. ~ftllc irenic is cubicctcd to e l a \ ~, tv loading alone, and if the frame geometry and loading are not so unsymmetrical as to cause significant 'sway' (lateral drift) of the frame, each floor may be malysed separately, considering the appropriate 'substitute frame', with the columns attached to the floor assumed fixed at their fay ends [Fig. 11.24(c)]. A further simplification may be made for thc purpose of determining the design moment at a given support or

'

It is now possible to do such analysis of the entire frame by methods such as the Finite Eler,te,rr Method. The successive levels of simplifications and approximations given below are for use when such computer-based methods we not resorted to, or ax found unnecessary.

/I

482

REINFORCED CONCRETE

DESIGN

DESIGN

span in the slab-beam member, by assuming the slab-bcam member to be fixed at any support two panels distant, provided the slab is continuous beyond this point [refer CI. 31..5.l(b) of theCode1. The load transfer system in the 'equivalent frame' involves three distinct interconnected ele~ncnts[Rg. 11.35(a)]: the slab-beam members (along span I , ) ; e the colutnns (or walls); and e the torsional members, transverse to the frame (along span 12) and along the column lines.

.

column above

(a) elements of equivalent frame at a connection

- w,h uer unit lenath

(b) equivalent frame for analysis

Fig. 11.35 Equivalent frame method

Tn conventional plane frames, the torsional members are absent, and the skeletal frame commises only beams and columns. However, in thc case of the 'equivalent frame', the wide slab-beam member is supported at its support section only over part of its width by the column, and the remaining (and generally substantial) portion is

OF TWO-WAY

SLAB SYSTEMS 483

supported by the transverse torsional member, which provides only elastic (flexible) restraint (spring support) - both rotationally (in a twisting mode) and translationally (in a vertical direction). In other words, significantly, tile flexural restraint to the slab-beam mcnlber (horizontal member) at the support section is less (i.e., the member is more flcn-iblc), than i n the case of a beam-column conncction in a conventional planc frame. In effect, it is as though the vertical supporting member (column) has less flcxural sliffness than it really has. The nonuniform translational restraint to the slab-beam member along a transverse line at the column is ignored in frame analysis, its effect being accounted for in the apportioning of the design 'positive' and 'negative' moments over the panel width to column strip and middle strips. However, the increased flexibility of the slab-tocolumi connection has to be accounted for in some manner in the assessment of the relative stiffnesses of the various members of thc 'equivalent frame'. Although the IS Codc provisions do not include any specific suggestion as to how this can be done, the ACI Code-(on which thc EFM procedurc of the IS Code is based) recommends the concept of an 'equivalent colomn' with stiffncss K,,which can be used to replacc the actual colunu~s(above and below the floor at any joint) as well as thc torsional mncmber at the colunrn line under consideration [Fig. 11.35(b)]. Thus, for the purposc of gravity load analysis, the substirim f i r m to he analysed by EFM can be modclled as a simple multi-bay, single storeyed portal fiamc, comprising only horizontal slab-beam members and vertical 'equivalent columns' [Fig. 11.35(b)J. The calculation of stiifnesscs of the slab-beam members and the equivalent colulmls arc to be based on their respective gross concrete sections [CI. 31.5.l(c) of the Codc]. Details of the calculation procedurc are discussed in the next section

I$11 jj

I1 :I

,,> ,!) I/

'1.1.1 I' I

I.

I

'!

'I

11.6.2 Slab-Beam Member The slab-beam member in an intedor frame is bot~ndedlaterally by the centrcline of the panel on each side of thc column line, thus comprising a column strip plus two half-middle strips. For an exterior frame, the slab-beam member extends laterally from the edge to the ccntreline of thc adjacent panel [Fig. 11.24(a)]. The slab-beam comprises the slab, drop panel (if provided) and beam(s) (if provided). The cross-section of the slab-beam member varies along its span, on account of provision of dmp . .panels (if provided) and the increased cross-section within the bounds of the supporting column; the consequent variation of second moment of area alonr- the span must be accounted for in the fratnc analysis . by . EFM ICI. 31.5.l(d)' . . of the Code]. In order lo account for the enhancement in the second nloment of area of the slab-beam member in tlic region between the column face and the colun~n centreline, a magnification Jacto~of (1 - cd12)' is reco~mnendcd[Ref. 11.18, 11.19]. The variation of tile second moment of area of thc slab-beam mcmber in a flat slab (with drop panels) is shown i n Fig. 11.36(a), (b), (c). 'With reference to waffle slabs (Yecessed' or 'cafferwY) which are mnrlc solid io the legion of h e columns [Fig. l.ll(b)l, the Code suggests that the stiffening elfect may be ignoied provided the solid par of Ihc slab does not extend more than 0.151, into the span measuremait from the centreline of the columns.

, j,

DESIGN OF TWO-WAY The calculation of the stiffness factors, carry-over factors and fixed-end moments of the slab-beam member (required for conventional frame analysis' ) are dependent on the variation of the second moment of area along the span. Such factors have been tabulated for common geometric and loading configuratiot~s in various design handbooks [Ref. 11.11, 11.25, 11.261. Factors for two typical cases are listed in Tables 11.5 and 11.6. The use of such tables is demonstrated in Example 11.7.

.Id). variation of I of column

(b) variation of lof slab

(SECTION 'CC'

(c) Sections through slab

Fig. 11.36 Variation of second moment of area along member axis

However, these factors are not required in modern computer-based analyses using the finite element method; nodes are introduced at the localians whem the second moment of area changes.

SLAB

SYSTEMS 485

Table 11.5 Moment distribution constants for slab-beam elements

486 REINFORCE0 CONCRETE

DESIGN

Table 11.5 (contd.)

Table 11.6 Moment distribution constants for slab-beam elements drop thickness = 0.50~7

DESIGN

488 REINFORCE0 CONCRETE DESIGN

Evidently, K., 2 ZK,: i.e., the effect of the flexibility of the torsional member is to reduce the rotational restraint offered to the slab-benm ~nemberat the support section. The condition K,. = ZK, (implying that the rotation along the entire length AB in Rg. 11.35(a) is tlle same) can be assumed to occur only if the torsional stiffness of the transverse torsional member is infinite or if the column is in fact a wall with a width extending ovcr thc full width of thc slab-bcam incmbcr. On the other hand, if ZK, = .-, K,, = K,, implying that although the column is infinitely stiff, the slab undergoes the same rotation as that of the torsional member along the length AB in Fig. 11.35(a) except for the width at the column location Another interesting result of K., = 0 is obtained for tlic hypothetical case of a torsional mcmbcr with K, = 0, or for the casc of a slab sitnply supported on a masonry wall (Kc = 0) rhroughoi~tthc length AB: In the former case, the slab is flexurally unrestrained along the length AB (except lor the width c2 at the column location), and in the latter case, the slab is flextirally unrestrained throughout the entire length AB, and in both cases K,, should naturally be zero. For the purpose of computing the torsional stiifness of the transverse member, the following approximate cxpresssion [Ref. 11.181is reco~nmeoded:

489

C z torsional property of the cross-section' [refer Eq. 11.411;and E width of the equivalcnt rectangular column, capital o r bracket, measured. transverse to the direction in which momcnts are being determined. . ... ~,. , it should beinote$,that the coQcoptoI $e.'eclu v8lent col~lmn~!jfJhessl::K&; ,o+plain 0.8

- OK.

4. Effective depths

*

Reinforcement will be placed in the outer layer for thc bars in thc N-S direction (in order to resist the larger moments in this direction), and in the inner layer for the bars in the G W direction. Assuming 16 $bars with ;I clear covcr of 20 mm, the effective depths arc obtained as: = 200 - 20 - 1612 = 172 mm &S I , , ,=, 300 - 20 - 1612 = 272 mm d,,,.,,,b = 172 - 16 = 156 mm ~ A L I Y ~=, 272 ~ ~ - 16 = 256 mm

5. Factored loads (i) self-weight of slab @ 25 x 0.2 = 5.0 kNlm2 = 2.7 " (ii) superimposed dead load (iii) live load = 2.5 " 10.2 kN11n' factored load on slab rv, = [M, - (V, x 0.25)+(122.4+ 11.25) x (0.25)~/2] !dim The values of VL, VR, x, M,f , M,,,and M,;

*

have been tabulated, using the

above formulas for thc various spans in Table 11.9' . As the slab satisfies the limitations for the Direct Design ~Mcthod,the sum of the d exceed 'positive' momcnt at midspan and average negative momcnt, M ,l ~ e not M,, the maximum static moment on the simply supportcd span I,, = 6.3 m [Fig. 11.56(f)]: Fig. 11.56 Factored moments

' These are indicated only for the first three spans, as the six-bay fimm is symmetric with

respect lo its middle.

in column and mtddle strtps in E-W d~rectlonExample 11.7

OESIGN

OF TWO-WAY SLAB SYSTEMS

523

The distributed moments in the column strips and half-middle strips in the various panels (in the E-W direction) are indicated in Fig. 11.56'.

9. Column moments The total unbalanced slab moments at the various supports are transmitted to the respective colunu~s. At each support, the unbalanced slab moment is shared by the column above and the column below in proportion to their relative stiffnesses.

7654 = 0.476 8.427 +7.654 with a carry-over factor = 0.595 (determined earlier) 0 Fraction of moment in column below = 1.0 - 0.476 = 0.524 with a carry-over factor = 0.54 The unbalanced slab moments are obtainable from Table 11.9: at eiterior column 1: 304 kNm at interior column 2: 641 - 583 = 58 lcNm at interior column 3: 531 - 519 = 12 k N n ~

a M,, = 0.08 [(I 1 . 5 5 + 0 . 5 ~ 3 . 7 5 ) ( 8 . 0 ~ 6 . 3 ~ 1.55x8.0~6.3')]/(1+ )-(I 113.456) = 0.08 x461.7 = 37 lcNm Accordingly, tllc unbalanced moment at the interior column 3 should be takcn as 37 !&ni (and not 12 kNm). The distribution of unbalanced moments to the various columns, above and below, including carry-over effects (using the coefficie~itsderived) are depicted in Fig. 11.57.

Fraction of moment in column above =

.

10. Flexural reiuforceme~~t The 'requirement of flexural (tension) reinforcement at all critical sections is computed in Table 11.10 for the columu strip (3400 rnm wide for 'positive' moments, and 3000 nunt wide for 'negative' moments), and in Table 11.11 for the middle strip (4600 nun wide for 'positive' moments and 5000 mm wide for 'negative' moments). 16 mm diametcr bars l~avcbccn used and the Code restrictions of minitnum reinforcement and maximum spacing have been adoptcd. Table 11.10 Design of reinforcement in column strip, E-W direction - Example 11.7

Fig. 11.57 Column moments (kNm)- Example 11.7 However, at interior column locations, the unbalanced moment should not be less than that given by Eq. 11.42: M,, = 0.08 [(w,,,,,+0.5w,,,~~)l~l,? - w : , ~ ~ ~ ;)']/(l+ (C

where w,,,, = w;,,

= 7.7

11%)

x 1.5 = 11.55 kNlm2 (neglecting drop panel)

= 2.5 x 1.5 = 3.75 kN/m2 1, =I; = 8.0 m, l,, =I,: = 6.3 m

W,,,LL

'

This distribution is nearly identical to one that will be obtained by following Ref. 11.18 and the median values of the rongr of factors given in Section 11.5.4 (sub-section (a)) for the Canadian Code.

It may be noted that, whcrc thc unbalanccd slab momcnt is significant (at the exterior support only, in this Example), adequate reinforcement should be providcd over a distance ~2 + 3D,l,o,,= 1400 mm, centred about the columri line, to permit the

'

The effective width of lhc colunu~strip at supports (for 'negative' moments) is restricted to that of the dmp pancl, for convenience.

524

DESIGN O F TWO-WAY SLAB SYSTEMS

REIN F O R C ED CO NC RETE D E S IG N

transfcr by flexure fiom the slab to the column thc poltion M,,eof the unbalanced moment. M,,e= 0.6 X 304 = 182 kNm

[Eq. 11.28aI 1 8 2 x i 0 6 R= brlZ 1 4 0 0 x 2 5 6 ~ 3 p, = 0.632 [Table A.3(a) or Eq. 5.12, for M 20 and Fe 4151 3 A,,=(0.6321100) x (1400 x 256) = 2265 mm" No. of 16 6 bars requil-cd = 22651201 = 12

- M,,6 =

alent frame analysis [Table 11.91, the maximum shear force, eqUal to , is found to be at the exterior face of first interior Support. Referring to 6(d), with* = 2.991 m and VR = 483 kN,and to Fig. 1 1 ' 3 5

= 256 mm from face of column, i.e., 506 m n from centre of Column, V,,{ = 483 - (122.4 + 11.25) x 0.506 = 415 id*T e shear force to be uniformly distributed over the width of the panel considering a 1 m strip of the slab outside the drop panel (where d =

=, ,

Thus, at tl~cexterior suppo~t,12 nos of 16 $bars lmlst be distributed over a width 1400 mm centred ovcr the column. The remaining outcr portions of the drop pan with width equal to (3000 - 1400)12 = 800 mm may be provided with two bars to limit the spacing to < 2 0 . The column strip 'negative' momcnt reinforcem tla exterior suppo~t,adds up to 12 + (2 x 2) = 16 bars, compared t o the 12 b indicated ill Tablc 11.10. 11. Check on one-rvAy shear stress l'hcrc arc two critical sections to be considcrcd: (i) r l 256 mm from thc face of column; and (ii) d = 156 mm from thc edge of drop panel. ;

(415~10')/~.~ ,333 Mpa 1000x156 which is less than kz, given by the Code (C1.40.2.1) for p, = 0.32 and M 20 (ii) st d = 156 mm from the edge of the drop panel, the shear force and hence the shear stress, will be less than that calculated above. This will obviously be safe.

k E o L ..-..-.. L.

..+ ........

Table 11.11 Design of reinforcement in middle strip, E-W direction - Example

,

.! 8

B;panei

T

....,......... :!

:i panel

::

!

I!

.:: . ....

..

! !

! ! r .

!

!

!

+..-

F I ~ 11.58 . Critical sections for one-way and two-way shear - Example 11.7

* minimom reinforcemei~t(A,, = 0.0012 6D)governs.

t maximum allowable spacing governs.

525

The applied shear stress z,., , therefore, marginally exceeds the shear strength T,, . Note, howcver that zV2< 1.5 iCz . Hence, shear reinforcen~entis required, with a total cross-sectional area:

12. Check on two-way (punching) shear stress The maximum column reaction and unbalanced slab moment (in an interio~ column) occurs at column 2 (or 5). From Table 11.9, the vertical reaction is obtained as the sum of the shears on either side as: R=483+442=925kN

a. Interior coluini~ The effective depth in punching shear calculations may be taken as the average of the effective depths in the E W and N S directions. The critical sections are (i) dl2 = (256 + 272114 = 132 mm from the column face all around; and (ii) dl2 = (156 + 172)14 = 82 lmn from the edge of the drop panel, as indicated in Fig. 11.58 i) F o r the critical section at dl2 from the column face, the punching shear is [Fig. 11.551 V,,z = 925 - (15.3

+ 3.75) x 0.764~= 914 kN

and the perimeter is

The moment transfemed by shear is 40 percent of the unbalanced moment M, = 58 kNm at colunum 2 [Eq. 11.28bl

.

Using 8 rmn $ stinups [Type I arrangement, Fig. 11,461, nurnber of vertical legs required on each side (of tllc square of side 764 nun) = 1528/(50.3 X 4) = 8. This can be achieved by providing 8-lcgged 8 nml$ stil~upscomprisi~lg4 nos 2-legged closed stimups [Fig. 11.461 on each side. Nominal 10 mm 4 holder bars may be provided at the stirrup corncrs if regular top steel and bottom steel are no1 otherwise available. This arrangement of stirrups should be provided at a spaciltg of not more than 0.75d = 0.75 x 264 = 198 1mn = 200 nun and sllould be continued to a distancc d = 264 mm beyond the section where the shcar strcss does not exceed 0.5 c,, . Checking at a section four spacings further removed from thc first crilical section. and now ignoring the marginal shear stress due to the unbalanced moment.

M,,, = 0.4 x 58 = 23.2 kNnt and the parameters c and J, in Eq. 11.51 [Fig. 11.45(a)J are given by: c = (c, + 412 = 76412 = 382 m n (c, + d ) d 3 J, = + (cI +d)3 d + (cl +d12(c2 + d ) d 6 6 2 764 x 264' 7643 x264 764' x 264 6 6 2 = 80.83 x lo9 nun4 Applying Eq. 11.51. +

- 914xlo3

+

(23.2x106)x382 3056 x 264 80.83 x lo9 = 1 . 1 3 3 + 0 l l O = 1.243MPa z,, = k,(0.25&) [Eq. 11.491

-OK.

+

= 1.0 ~ 0 . 2 5 =6 1.118 M P ~ '

'

;:G

:

-

Unsupported L e n g t h The Code (CI. 25.1.3) defines the 'unsupported length' 1 of a colunm explicitly for various types of constructions. In conventional framed constroction, 1 is to be taken as, the clear dismnce between the floor and the shallower bensr fruming into tire olumns in each direction at the next higher floor. level. By tlus, it is implied that en a column is framed in any direction by beams of diffcrcnt depths on either side, n the u~isupportedlcngth (with respect to buckling about a perpendicular axis) 'For a rectangular scction. r

-

030: for a circular section, r = 0.25D. wen there exists r8ative translation of the ends of the column member, the points uf flectian (zcro moment) may not lie within the membcr. In such cares. they may be located y extending the dcnection curve beyond the calunm end($) and by applying conditiuns of symmetry, as shown in [Fig. 13.4(a), (b)l.

570

DESIGN

R EI N F O R C ED CONCRETE DESIGN

shall be considered, conservatively, with reference to the shallower beam. It may be noted that the unsupported length in one direction may be different from that in t perpendicular direction. For a rectangular column section (width D,x depth D,) m a y w e the terms, 1, = k, 1, and I , = k,. 1, to denote thceffectivc lengths rcfe~rin buckling about the major and minor axes respectively, where 1, and 1, denote corresponding unsupported lengths and k, and k, denote the corresponding effe length factors. These concepts are made clear in Fig. 13.2a, and further illustrate Examples 13.1 and 13.2.

X

Major axis iY

Minor axis

.

i

OF

COMPRESSION MEMBERS

571

stiffness), and the othcr extreme value k = 1.0 corresponds to zero rotational fixity at both colulnn ends ('pinned') [Fig. 13.3(c)] (i.e., when the beams have zero flexural stiffness). When one end is fully 'fixed' and the other 'pinned', k = 0.7 [Fig. 13(b)].

(4 plan '7,s 0,

1" s &

slenderness ratios:

(a)

both ends rotationally fixed

top of lower floor

(c)

both ends rotationally free

(d)

both ends partiall) restrained (rotational)

Fig. 13.3 Effective lengths of columns bracedagainst sideway

(b) section at X.

Flg. 13.2a Definitions of unsupported and effectivelengths in a rectangular column

In the case of 'flat slab construction', the unsupported length 1 is to be taken as the clear. distance between the floor and the lower extremity of the capital, the drop panel or slab, whichever is the least. '

(b)

one end rotationally fixed, the other free

13.2.2 Effective Length Ratlos f o r ldealised B o u n d a r y Conditions When relative transverse displacement between the upper and lower ends of a column is prevented, the frame is said to be braced (against sidcway). In such cases, the effective length ratio k varies betwecn 0.5 and 1.0, as shown in Fig. 13.3. The extreme value k = 0.5 corresponds to 100 percent rotational fixity at both column ends [Fig. 13.3(a)l (i.e., when the connecting floor beams have infinite flexural

When relative transverse displacement between the upper and lower ends of a column is not prevented, the frame is said to be urrbmced (against sideway). In such cases, the effective Icngth ratio k varies between 1.0 and infinity, as shown in Fig. 13.4. The lower limit k = 1.0 corresponds to 100 percent rotational fixity at both column ends [Fig. 13.4(a)], and the upper theoretical k = = corresponds to zero rotational fixity at both column ends, i.e. a column pinned at both ends and permitted to sway (unstable) [Fig. 13.4(c)]. When one end isfully 'fixed' and the other 'free', the column acts like a vertical cantilever in the buckled mode, con'esponding to which k = 2 [Fig. 13.4(b)].

DESIGN

572 REINFORCED CONCRETE DESIGN

IP

OF

COMPRESSION

b) one end 'fixed' and the other 'partially fixed' c) one end 'fined' and the other i.ee [ ~ i g 13.4(b)l .

MEMBERS

573

: 1.50 : 2.00

The most common casc encountered in framed buildings is thc one involving

ed' , it is desirable to assume a more conservative estimate - s a y k = 1.0 or e. However, if the frame is clearly 'unbraced', it is necessary to ascertain thc

13.2.3 Effective Length Ratios of Columns in Frames tational restraint at a column end in a building frame is governed by the xural stiffnesses of the members framing into it, relative to the flexural stiffness of

P

(C)

both ends rotatlonallv free

(a)

both ends rotationallv fixed (b)

(d)

both ends varllallv restrained (rotational)

one end rotationally fixed, the other free Fig. 13.4 Effective lengths of columns unbracedagainst sideway

Code Recommendations for ldealised Boundary Conditions Although, in design practice, it is convenient to assume the idealised bound conditions of either zero or full restraint (rotational and translational) at a colu the fact is that such idealisations cannot generally be realised in actual struc this reason, the Code (CI. E-I), while permitting these idcalisations, recom use of 'effective length ratios' k = ;,I1 that are generally more conservative obtained from theoretical considerations [Fig. 13.3, 13.41. These recommended values of k are as follows: 1. coLmms braced ngninrr sidewny: a) both ends 'fixcd' rotationally [Fig. 13.3(a)] b) one end 'fixed' and thc other 'pinned'[Fig. 13.3(b)] C ) both cnds 'free' rotationally ('pinned') [Fig. 13.3(c)l

: 0.80 (instead of : 1 .OO

2 . col,rnms u~zbmcerlngninst sideway:

a) both ends 'fixed' rotationally [Fig. 13.4(a)l

: 1.20 (instead of 1.

The IS Code (CI. E-I) recolmnendations are based on design charts proposed by Wood [Ref. 13.31. The Bdtish Code [Ref. 13.21 and the Commentary to the ACI Code [Ref 13.11 recommend the use qf certain simplified formulas, which are ~larlysuitable for computer-aided design. Other methods, including the use of 'alignment charts' [Ref. 13.4, 13.51, have also been proposed. All of these ds provide two different sets of chartslformulas: one set for columns 'braced' t sideway, and the other set for 'unbraced' colunms. This is a shortcoming in

-

'unbraced'. and the difference evented from side-swavl .. and rarelv comvletelv . etween the two estimates of effective length ratio csn be considerable. A recent tudy [Ref. 13.221 shows how this problem can bc resolved using fuzzy logic oncepts, which incorporates the concept of 'partial bracing'. This aspect of 'partial ng' may be more accurately accounted for by means of a proper second-order is of the entire frame is rcquired [Ref. 13.61; howcvcr, this is cornputationally ' t for routine design problems.

Whether a Column Is Braced or Unbraced

-

roximate way of deciding whether a column is 'braced' or 'unbraced' is given CI Code commentary. For this purpose, the 'stability index' Q of a storey in a torcyed building is defined as:

.ally, the assumption of a fully braced frame can be safely made if there rue special elements in a building such as shear walls, shew trusses, etc. [Ref. 13.71 (which are to resist practically all the lateral loads on the frame). Even otherwise, if a rigid ssesses sufficient inherent translational stiffness, and especially if there mc in-fill walls, it may be considered to be braced - at least pxtially, if not fully.

574 REINFORCED CONCRETE DESIGN

where

c, =sum of axial loads on all columns in t l ~ estomy;

h, = height of thestorey; A,, -elastic first-order lateral deflection of the storey; H,, -total lateral force acting on thc storey. It can be shown [Ref. 13.81 that, in the absence of bracing elements, the 'lateral flexibility' measllre of the storey A J H , (storey drift per unit storey skiex) may be taken (for a typical intermediate storey) as:

j C ? !'

: i '

where 21,-sum of second moments of areas of all columns in the storey in the

The application of this concept is demonstrated in Example 13.1.

0'7+0'05(a1 0.85+0.05 a ,,,,,,

is less, for braced columns

L'0+0'15(a1 2.0+0.30 a,,,,,,

whichever is less, for unbraced columns (13.5b)

(13.W 8

For a fully fixed condition, a = 0 may be considered, and lor a 'hinged' condition,

Use of Code Charts

a = 10 may be considered.

xI

~

~h~ followillg fortllulas, givcn in BS 8110 [Ref. 13.21 and the CoJluncntarY to the ACI code [ ~ ~13.11, f . providc useful estimates of the effective length ratio k:

the storey in the plane under consideration; E, = modulus of elasticity of concrete. Eq. 13.2 is derived by assuming points of inflection at the mid-heights of all columns and midspan locations of all beams, and by applying the unir load method to an isolated storey [Ref. 13.81. If special bracing elements such as shear walls, shear trusses and infill walls are present, then their effect will be to reduce A /?I .. -XI--,, significantly. l'

Charts are given in Fig. 26 and Fig. 27 of the Code for determining the effective length ratios of braced columns and unbraced columns respectively, in terrns of coefficients p, and p, which represent the degrees of rotational freedom at the top and bottom ends of the column [Ref. 13.71:

I?,

Use of F o r m u l a s

plane under consideration; - 16/16 -sum of ratios of second moment of area to span of all flodr members in

'

whcre the notation jr denotes that the summation is to be done for the members framing into the top joint (in case of PI) or the bottom joint (in case of P2). T h e increased beam stiffness for unbraced columns [Eq. 13.4b1, c o ~ ~ ~ p a rtoe dbraced columns [Eq. 13.4~11, is attributable to the fact that in the case of the latter, the beams are bent in single curvature, whereas in the case of the former, the beams are bent i n ... . " double curvature, in the buckled configuration. The limiting values P = 0 and P = 1 represent 'fully fixed' and 'fillly hinged' conditions respectively. The use of these Codc charts (not reproduced in this book) is demonstrated in Examples 13 1 and 13.2.

The application of these formulas is den~onstratedin Exa~~lples 13.1 and 13.2

:

EXAMPLE 13.1

The framing plan of a multi-storeyed building is shown in Fig. 13.5(a). Assume that all thc columns have a size 300 nun x 400 nun; the longitudi~lalb e a m (global Xdirection) have a size 250 n ~ mx 600 nun and thc transverse beams (global Ydirection) have a size 250 nun x 400 nun as shown. The storey heigl~th, = 3.5 in. For a column in a typical lower floor of the building, determine thc effective lengths 1... and 1," with respect to Lhc local x- and y - axes (major and minor), as shown in Fig. 13.5(b). For the purpose of estilnati~~g the total axial loads on the columns i n the storey, assume a total distributed load of 35 l c ~ 1 1 2from all the floors above (combined). Also assume M 25 gradc concrete for the columns and M 20 grade concrete for the

11:

576

REINFORCED CONCRETE

DESIGN

DESIGN

3

.

C I , / h , =16x

OF

COMPRESSION MEMBERS 577

300x4003112 = 7314x10'mm3 (for sway in the global Y3500

direction). Longitudinal Bcams: 12nos. 250 m m

X

600 nun,I,, = 6000 Ilml

-3 * ~ I , , / I ~=)1 2~x (250)~(600)'/l2 = gOOO 6000 Transverse Beams: 12nos, 250 nun X 400 mm,1, = 4000 mm (250)~(400)'/12 = 4000 -3 3 ~ I , , / I , =12x ,)~ 4000

Columns Braced o r Unbraced ? Lnteml Flexibility measures of the storey: (A,, /H,,). and (A,, / H , , ) ~ Ignoring the contribution of in-fill walls [Eq. 13.31:

TYPICAL FRAMING PLAN

where E,

(a)

=5

0 0 0 K (as pcr Code C1.6.2.3.1)

For columns, fck = 25 MPa 3 E , , , = 5000zJi;;

= 25000 MPa

For beams, fck = 20 MPa =, E,,,

= 22361 MPa

= 5000fi

Longitudinal direction (global X-dkection): 3 h..

'Y LOCAL AXES OF COLUMN

SECTION A - A

.

= 1.4998 x 10.' mmJN Transverse direction (global Y-direction)::

(b)

Fig. 13.5 Example 13.1 SOLUTION

Unsupported lengths of c o l u ~ m 1,= 3500 - 600 = 2900 mm (for buckling about y-axis) 1, = 3500 - 400 = 3100 nun (For buckling about x-axis) Relative sriffnness memares of columns and b e a m Columns: Ibnos, 3001nm x 400 nun, h, = 3500 m m =, C I , / h , =16x 400x300"12

3500

direction), and

= 4114x10'mm3 (for sway in the globalX-

Stability Index Q Total axial load on all columns = 35 kN/mZ x (12.0 m x 18.0 m) = 7560 kN

Longitudinal direction: Qu = 7560 In' x 1.4998 x lo-' = 0.0324 3500 7560 'lo' x1.6996 x Transverse direclion: Q y = = 0.0367 3500

As Qx = 0.0324 < 0.05. the storey can be considered 'braced' in the longitodiaal direction. As QY= 0.0367 < 0.05, the storey can be collsidered 'braced' in the transverse direction. Hence, the columns in the storey may be assumcd to be 'braced' in both directions. (Note that Qx and Q Y are less than the IS Code limit of 0.04 as well)

0.7 +0.05(2x0.3427) = 0.7343 (lesser) 0.85+0.05(0.3427) = 0.8671 3 1, = 0.7343 x 2900 = 2129 mil

* k, =

t

914 with resncct to moio,- (loco1 x-1 axir: a, = a 2=- = 1.3703 667

-p

Effective Lengths by IS Code charts 3

Clclhs,

PI=A=

If

[ ~ q 13.4a] .

C1clhs+ ~ 0 ~ 5 ( 1 6 / 1 b ) b

Bucklinr with r e s ~ e ctof minor (local v-J axis:

Note 2: Alternatively, the designer may assume idealised boundary conditions braced columns with partial rotational fixity at top and bottom. Assuming a valuc k = 0.85 (as explained in Scction 13.2.2) =, 1, = 0.85 x 2900 = 2465 I-, and 1, = 0.85 x 3100 = 2635 mm This results in a slightly conservative estimate of effective length

Z ( l c / h r ) = 400x3003112x2=514x10~n11,,3 B 3500

Notc 3: A realistic assessment of effective length is called for in the case of slerzder colu~ms. In the present case, as 1, /D, and I , ID, are approximately 7, nnd well below 12, the column is definitely a short column, and there is no real need for a rigorous calculatio~lof effective

* /I,.

e

{

Note 1: The use of formulas gives effective lengths that are generally within t 8 percent of the corresponding values obtained frwn the Code charts.

Jt

Referring to Fig. 26 of the Code, k, = 0.64 = 1, X k, = 0.64 x 2900 = 1856 Bucklina with resuecr to major (local x-) axis:

0.7 +0.05(2x1.3703) = 0.8370 (lesser) 0.85+0.05(1.3703) =0.9185 1, = 0.8370 x 3100 = 2595 mnl k, =

,.

.i l i

EXAMPLE 13.2 -

'! , ,,

Repeat the problem in Example 13.1, considering a column size of 250 1nln x 250 nun (instead of 300 mm x 400 mm). SOLUTION

Referring to Fig. 26 of the Code, k, = 0.82 1, = 1, X k, = 0.82 x 3100 = = 2542 nlm

.

*

Longitudinal Beams: ( ~ I ~ , / =/ 9000 ~ ) ~x 10' mm3(as in Example 13.1)

Alternative: Effective Lengths by Formulas [Eq. 13.5a]

x

1,/J?,

Unsupported lengths of calumn ly = 2900 mil and I, = 3100 nun (as inExample 13.1) Relative stiffness r~~casrrres of columns and beams Colu~mls:z l- , / h , =I6 x (250)'/(12 x 3500) = 1488 X lo3 mn13

.

Transverse Beams:

(21,/1,)~ = 4000 x lo3nun3 (as in Example 13.1)

Lateral Flexibilily ~ncasa,r.cof the storey:

~ u b s t i t u t i nE,,,! ~ = 25000 MPa, EC$L,,, = 22361 MPa (as in Examplc 13.1) and h, = 3500 mm, and values of relative stiffness measures,

. ?$ 3

,,I,

I

. ' ,. ,.,

\

580

DESIGN O F COMPRESSION ME MB ER S

REINFORCEDCO NC RE T E DESIGN

581

Alternative: Effective Lengths by Formulas [Eq. 13.5al Longitudinal direction (globalx-direction): 2

=

3.2514 x 10" mm/N

)x Transverse direction (global Y-direction):

[k Iy -"-

= 3.8855 x 10" mm/N

[Compwing these values with those obtained in Example 13.1, it is seen that the reduction in column size rcsults in a drastic increase (mote than double) in the lateral flexibility of the storey]. Stability Index Q P,, = 7560 !dV (as in Example 13.1)

1.0+0.i5(2x0.124) = 1.0372 (lesser) 2.0+0.30(0.124) = 2.0372 g 1, = 1.037 x 2900 = 3007 mm r Btlcklina with respect to maiof~(localx - ) axis:

e x = 7560 x103 ~ 3 . 2 5 1 4x ~ o =0.0702>0.05 - ~ 3500

= 7560

x 3.8855 x 10-I = 0.0839 > 0.05 3500 Hence, the columns in the storey should be considcrcd as 'unbraced' in both directions. QY

to rnlnor (local v-I axis: a, =a, =-186 = 0.124 1500

1.0+0.15(2~0.2789)=1.084 2.0+ 0.30(0.2789) = 2.084

186 667

a, =a, = - = 0.2789

(lesser)

~l,=1.084x3100=3360mm Note 1: The effective lengths predicted by the two different methods are fairly

Effective Lengths by IS Code charts

C I, lhS

PI = Pz =

zlc/h$ 11

Note 2: Considering the effective lengths given by the Code charts, the slenderness ratios of the column are obtained as follows: I,JD,= 30161250 = 12.1. = 33791250 = 13.5; The column should be designed as a 'slender column'.

[Eq. 13.4 (b)]

+C1,5(16/l6)

jr

jr

Bucklinn wirh resnect to minor (local v-) axis: z(lc/hs)=250'112x2 3500 b

= 1 8 6 x 10~mm'; x(1,/16) = 1500x 10'mm3 b

13.3 CODE REQUrREMENTS ON SLENDERNESS LIMITS, MINIMUM ECCENTRICITIES AND REINFORCEMENT

(as in Example 13.1), 186 PI = 0 2 = 186+1.5(1500) = 0.076 Referring to Fig. 27 of the Code, k, = 1.04 ~ l , = l , x k , = 1.04x2900=3016mrn Buckline with rrsnect fo nraior (localx- j axis: ~ ( I , / h , ) = 186 x lo3 mm3 ; z ( I , / l , ) 11

PI = P 2 =

=0.157 186+1.5(667) Refel~ingto Fig. 27 01the Code, k, = 1.09 *l,=l,xk,= 1.09x3100=3379mm

= 667 x lo1 nun3(as in Example 13.1)

13.3.1 Slenderness Limits Slenderness effects iq columns effectively result in reduced strength, on account of the additional 'secondary' moments introduced [refer Section 13.71. In the case of very slender columns, failure may occur suddenly under small loads due to instability ('elastic buckling'), rather than due to material failure. The Code attempts to prevent this type of failure (due to instability) by specifying certain 'slenderness limits' in the proportioning of columns. The Code (Cl. 25.3.1) specifies that the ratio of the unsupported length (1) to the least lateral dimension (d)of a column should not exceed' a value of 60: I/d < 60 (13.6) Furthermore, in case one end of a column is free (is., cantilevered column) in any given plane, the Code (CI. 25.3.2) specifies that

@)

' In the case of 'unbraced' columns, it is desirable to adopt a more stringent limit - say.

DESIGN

* 15 1 0 0 b Z / ~

(13.7) where D is the depth of the cross-section measured in the plane of the canlilcver and b is the width (111 the perpendicular direction).

*

As explained in Section 13.1.2, the general case of loading on acomprcssion tncmbcr is one comprising axial compression con~binedwith biaxial bending. This loading condition is represented by a state of b i a i u l eccentric cantpression, wherein the axial load P acts eccentric to the longitudinal centroidal axis of the colunm cross-section, with eccet~triciticse., and e, with respect to the major and minor principal axes .(Fir. 13.21c)l. . .. Very often, eccentricities not explicitly arising out of structural analysis calculations act on the column due to various reasons, such as: lateral loads not cottsidered in design; live load placements not considered in dcsign; accidental lateralleccentric loads; errors in construction (such as misalignments); and slenderness effects underestimated in design.

For this reason, the Code (CI. 25.4) ~.equiresevery column to be designed for a minimum eccentricity en,, (in any plane) equal to the unsupported lengt11/500 plus lateral dimensiod30, subject to a minimum of 20mm. For a colu~nn with a rectangular section [Fig. 13.21, this implies: ex.""m =

20 mm l/500+ D,/30

(whichever is greater)

(13,Ba)

(whichever is greater)

(13.8b)

1,/300 20 Inn1

(whichever is greater)

to ensure notninal flexural resistance under unforeseen eccentricities in loading; and to prevent the yielding of the bars due to creept and shrinkage effccts, which result in a transfa of load f r o ~ nthe concrete to the steel.

Morinrtan Keirrforcemozr: The maximum cross-sectional area of longitudinal bars should not excccd 6 percent of the gross area of the column section. However, a reduced maximum limit of 4 percent is nxonunended in general in the interest of better placcrnent and compaction of concrete -and, in particular, at lapped splice locations. In tall buildings, colunms located in thc lowernlost stoveys gencially carry heavy rcinforcement (- 4 percent). The bars are progressively curtailed in stages at higher Levels.

For n o ~ ~ e c t a n g u l aand r non-circular cross-sectional shapes, it is recommended [Ref. 13.71 that, for any given plane, e twn. = {

MEMBERS 583

h~ very large-sized columns (where the large s i x is dictated, for instance, by al-chitecmral consideralions, and hot strength) under axial compression, the limit of 0.8 percent of gross area may rcsult in exccssive reinforccmcnt. In such cases, the Code allow some concession by pernlitung the minimum arca of steel to be calculated as 0.8 percent of the areu of concrete required to resist the direct stress, and rtot the actual (gloss) area. However, in the case ofpedestals (i.e., compression members with 1dD < 3) which arc designed as plain concrete columns, the minimum requiretncnt of longitudinal b a s may bc taken as 0.15 percent of the gmss area of cross-section. In the case of reinforced concrcte walls, the Code (C1. 12.5) has iutroduced detailed provisions regarding minimum reinforcement requirements for vcrtical (and horizotltal) stccl. The vertical reinforcement should not be less than 0.15 perccnt of the gross area in general. This may be reduced to 0.12 perccnt if welded wire fabric or deformed bars (Fe 415 / Fe 500 grade steel) is used. providcd the bar diamctcr does not exceed 16 nun This rcinforcement should bc placed in two layers if thc wall is more than 200nun thick. In all cases, thc bar spacing should not excced three times the wall tltickness or 450 nun, whichever is less.

13.3.2 Minimum Eccentricities

s

OF COMPRESSION

Mininzum dinmeto. / ,zro,tbcr. of 60,s and their location: Longitudinal bars in columns (and pedestals) should not be less than 12 111111 in diameter and should not be spaced more than 300 ,nun apart (centre-to-centre) along the periphery of the colunu~'[Fig. 13.6(a)]. Al least 4 bars (one at each comer) should be provided in a colunui with rectangular cross-section, and a1 lcast 6 bars (cqoelly spaced near the periphery) in a circular colunln. In 'spiral columns' (including noncircular shapes), the longitodin;d bars should bc placed in contact with the spiral

(13.8~)

where 1, is the efective length of the column in the plane considered

13.3.3 C o d e R e q u i r e m e n t s o n R e i n f o r c e m e n t a n d Detailing Longitudinal Reinforcement (refer Ci. 26.5.3.1 of the Code)

*

Minimum Reinforcement: The longitudinal bars must, in general, have a crosssectional area not less than 0.8 percent of the gross area of the column section. Such a minimum limit is specified by the Code:

'

Cltep effects can be quitc lpro~~ouncerl in compression members undcr sustaincil sel-\,ice loads [refer Section 13.4.2]. l'hc consequeut increase i n steel slrcss (due to creep stcailt) is found lo be relatively high iat very low reinforcetnent percentages: hence. Ole mij~i~mtrn limit of 0.8 percent ispaescribed [Ref. 13.71. In the case ol reinforced concrae wdls, the Code (CI. 32.5b) ~.ecom~noaris 8 manmum spacing of tliree times the wall thickness or 450 mm. whichever is smaller.

'$,

584

REINFORCED CONCRETE

DESIGN

DESIGN

reinforcement, and equidistant around its inner circum~crence[Fig. 13.6(b)l. In columns with T-, L-, or other cross-sectional shnpcs, a1 least one bar should be locatcd at each corner or apex [Fig. 13.6(c)l. Longitudinal bars are usually located close to the periphery (for better flexural resistance), but may be placed in the interior of thc column when eccentricities in loading are minimal. When a large number of bars need to be accommodated, they may be bundled, or, alternatively, grouped, as shown in [Fig. 13.6(d)l.

OF

COMPRESSION

MEMBERS

585

desirable, in the interest of durability, to provide increased cover [Table 5.11 bot preferablynot greater than 75 mm.

Transverse Reinforcement (refer CI. 26.5.3.2 of the Code) General: All longitudinal reinforcement in a compression member must be enclosed within transverse reinforccment, comprising either lateral ties (with internal angles not exceeding 135') or spbuls. This is required:

* * * *

to prevent the premature buckling of individual bars; to confine the concrete in the 'core', thus improving ductility and strength; to hold the longitudinal bars in position during construction; and to provide resistance against shear and torsion, if required.

Lateral Ties: The arrangement of lateral ties should be effective in fulfilling the above requirements. They should provide adequate lateral support to each longitudinal bar, thereby preventing the outward movement of the bar. The diameter of the tie @, is governed by requirements of stiffness, rather than strength, and so is indepcndcnt of the grade of steel [Ref. 13.71. The pitch s, (centre-tocentre spacing along the longitudinal axis of the column) of the ties should be small enough to reduce adequately the unsupported length (and hence, slenderness ratio) of each longitudinal bar. The Code recommendations (based on Ref. 13.9) are as follows:

(13.10) where $,o,, denotes the diameter of longitudinal bar to be tied and D denotes the least lateral dimension of the column. Ideally, the tie must turn around (and thereby provide full lateral restraint to)

Fig. 13.6 Some Code recommendations for detailing in columns

vided for the corner and alternate bars [Fig. l3.6(e)l. The straight portion of a sed tie (between the comer bars) is not really effective if it is large, as it tends to ge outwards when the concrete core is subjected to compression [Ref. 13.10]. r this reason, supplenlentary cross ties are required for effective confinement of

The ends of every tie (whether closed or open) should be properly anchored. In case of grouping of longitudinal bars at the corners of a large-sized column greater) is specified for walls.

However, in aggressive environments,

588 REINFORCED

'.

CONCRETE

DESIGN

DESIGN

1. the amount of creep, which is influenced by the history of sustained loading and numerous factors related to the quality of concrete [rcfcr Section 2.1 11. 2. the amount of shrinkage, which in turn dcpcnded on the age of concrete, method of curing, environmental conditions and several other factors related to the quality of concrete [refer Section 2.121 In general, the strain in the cross-section E, increases with age on account of Cree and shrinkage, with a consequent redistribution of stresscs in concrete and steel, sltc that thc load shared by the concrete is partially transfen-cd to the steel. Consequent$, it becomes difficult to predict the stresses f, and f,, (in Eq. 13.12) under servic loads. According to conveutional working stress method of design, substituting the per.rrtissible sr,rsses ac,and a,, in lieu off, atidf,, rcspcctively, the design equation is obtained lrom Eq. 13.12 as

OF

COMPRESSION MEMBERS 589

:However, beyond the ultimate load (point B in Fig. 13.8). the behaviour depends on the type and amount of transverse reinforcement.

Tled columns Generally, the longitudinal steel would have reached 'yield' conditions at the ultimate toad level P, [point B in Fig. 13.81 -regardless of whether transverse reinforcement is provided or not'. However, in the absence of transverse reinforcement (or with widely spaced lateral ties), failure will be sudden atid brittle, caused by crushing and shearing of the concrete (as i n a plain concrete cylinder test - refer Section 2.8) and accompanied by the buckling of longitudinal bars. In the case of tied columns, some marginal ductility [paths BC, BD in Fig. 13.81 can be introduced by providing closely spaced lateral ties which undergo yielding in tension prior to collapse of the colum~s. The descent in the load-axial shortening curve is attributable to 'softening' and micro,c~ackingin the concrete. , .., '!!

where as,is takcn approximately as 1.5maC,(as in doubly reinforced beanis - ref Section 4.6). However, this assumption renders the steehtress as,independent o grade of steel, and results in unrealistic and uneconomical designs. The Code (B in its provision lor working stress design, attempts to somewhat remedy this situ by recommending

1:.

.,

t

1

130MPa for Fe 250steel 190MPa forFe415,Fe500stcels The allowable stresses in concrete (acc) under direct compression are specified as Ox'=

I

4.0MPa 5.0MPa o,, = 6.0MPa 8.0MPa 9.0MPa

h r M I5

M 20 for M 25 for M 30 for M 35

for

However, most codes of other countries have dispcnsed with the working s method (WSM) of design altogether, with the advcnt of the rrltimare load (ULM) of dcsign initially, and the more rational liririr srates method (LSM) of subsequently (since the 1980s). Indeed, in the revised Indian Code too, p givco to the LSM design procedure and the WSM relegated to an Atincx.

13.4.3 Behaviour Under Ultimate L o a d s Unlike service load conditions, the behavioltr 01an axially con~prcssedshort co' is fairly prcdictable under ultimate load conditions. It is found that thc strength of the columm is relatively independent of its age and history of 10. axial loading is increased, axial shortening 01 thc colulml increases line. about 80 percent of the ultimate load P,,, (path OA in Fig. 13.8); this be found to be independcnt of the type of transverse reinforcement [

axial shortening

.

Fig. 13.8 Behaviour of axially loaded tied and spiral columns

'th the spiral colunm that substantial ductility is achieved prior to the collapse olumn [path BE in Fig. 13.81. It is found that, approximately at load level P,,, oint B in Fig. 13.81, the outer shell of the concrete (covering the spiral) spalls off; t he concrete in the 'core', laterally confined by the helical reinforcement, ues to can'y load Collapse ultimately takes place when the spiral reinforcement in tension. The load can'ying capacity after the spalling can exceed P,,o v~dedthe amount of spiral reinforcement is such that the load capacity contributed it more than makes up for the loss in load capacity due to spalling 01the concrctc be noted, however, that, for design porposes, the Code limits the ultimate strain i n to 0.002, as a conservative measure. Corresponding to this strain, yield condilioos ot be attained in the case of Fe 415 and Fe 500 grades of steel [refer Fig. 3.6,3.7].

DESIGN OF COMPRESSION MEMBERS 593

592 REINFORCED CONCRETE DESIGN

Design of LongififrlinolReinforceme111 =0.4fekA8 +(0.67fy-0.4fck)A,, [Eq. 13.171 +, 3000 x lo3 = 0.4 x 20 x (450 x 600) + (0.67 x 415-0.4 x 20)A, = 2160x10' + 270.O5Ax -A,, = (3000-2160) x 10~1270.05= 3111 mm2 In view oC the column dimensions (450 mm, 600 nini), it is necessary to pla intermediate bars, in addition to the 4 corncr bars: Providc 4-25 $ a t corners : 4 x 491 = 1964 mm2 and 4-20 $additional: 4 x 314 = 1256 mm2 A,, = 3220 inm2 > 3111 mml

Minimum eccerrtricity

c,

a y = (100x3220) I(450x600) = 1.192 > 0.8 (minimum reinf.) -OK. 4-25

rn (at corners)

8 4 TIES @ 300 CIC

P,, = 1500 kN (given) = 1.05 [0.4fckAA,+ (0.67& - 0 . 4 L J for spiral columns (appropriately reinforced) esign of longifudinrrl reinforcenrent:

=, 1428.6 x 10' = 1256.6 x lo3+268.05 A,, 3 A,, = (1428.6 - 1256.6) x 103/268.05

= 642 rmn2 (equal to 0.51% of gross area). A,,,,,,;,,at 0.8% of A,

provide 6 nos 16 $:A, = 201 x 6 = 1206 I& esign of Spirul reirtforcernenf Assuming a clear cover of 40 mm over spirals, Core diameter = 400 - (40 x 2) = 320 nun Assuming a bar dianleter of 6 lmn and pitch s, Volumeof spiral reinforcement per unit Volumeof core (n x 6'14) x rr x (320 - 6)/s, -

Fig. 13.9 Example 13.3 Lrrrcrul Ties Tie diamctcr $, >

y compressed short coluinns may bc used.

: provide 8 nu11 din;

> 1005 mm2.

of colu,,,,l

0.3468

: pmvide 300 mm.

ent (Eq. 13.15, CI. 39.4.1 of Code) :.Provide 8 $ties @ 300 c/c The detailing of reinforcement is shown in Fig. 13.9

Design the reinforcement in a spiral column oC 400 nxn diameter subjected to a factored load of 1500 kiV The column has an unsupported length of 3.4 m an braced against sideway. Use M 25 concrete and Fe 415 steel. SOLUTION Sltorr Colunttr or- Slender Colfrrm ? Given: I = 34?0 nun, U = 400 mm =, slenderness ratio = 1,lD (as column is bmced). As 1JD l~ 12, the column may be designed as a slror.1colarrrn.

.

< 34001400 =

core dial6 = 53.3 mm

594 REINFORCED CONCRETE DESIGN

DESIGN 01; COMPRESSION

MEMBERS 5s5

Provide 6 $spiral @ 28 mnm clc pitch The detailing of reinforcement is shown in Fig. 13.10. 6 $ spiral

B 28 c/c pitch (clear cover = 40 mrn

n, and

tensile straills on ollc side of the NA and compressive strains on the

6-16$

with the maximum strain in the highly linearly varying across the pressed edge, &,, havillg a value between 0.002 and 0.0035 at the ultimate limit tate. This is depicted in the Fig. 13.1 I .

13.5 DESIGN OF SHORT COLUMNS UNDER COMPRESSION WITH UNIAXIAL BENDING This section deals with the behaviour and design of short compression members subject to axial compression combined with uniaxial bending, i.e., bending with respect to either the major axis or minor axis (but not both). AS explained in section 13.1.2, this loading condition is statically equivalent to a condition of uniaxial eccentric compression wherein the factored axial load P,,is applied at an eccentricity e =M,,IP,, with respect to the centroidal axis, M,, being the factored bending The traditional 'workkg stress method' of design is not covered in this section, not only because of the fact that it has become obsolete, but also because the code (C1. B 4.3) makes itmandatory that designs for eccentric compression by WSM, based on 'cracked section' analysis' should be further checked for their under ulrirnate load conditions to ensure the desired margin of saf~ry. hi^ condition effectively makes WSM redundant, as it suffices to design in accordance with LSM. 13.5.1 Distribution of Strains at Ultimate Limit State A special limiting case of uniaxial eccentric compression is the conditiorl of zero eccentricity ( e = 0, i.e.. M,, = 0) which corresponds to the axial loading condition,

discussed in Section 13.4. Col~espondingto this condition, the strain across the colurm section is uniform and limited to E, = 0.002 at the h i t state of collapse in comprrssion (as per the Code).

'

'Uncracked section' analysis is pernutted by the Code (C1. 46.1) when the eccentricity ir loading is so slnall that the resulting flexural tension, if any, can be borne by the concrete.

CROSS SECTCON

Fig, 13.1, possible strain profiles

under ultimate iimit state ineccentric compression

DESIGN OF

596 REINFORCED CONCRETE DESIGN

It may be noted that all the assumptions made in the analysis of the ultimate li state in flexure [[refer Section 4.71 -excluding thc one related to the minimum ten strain E~,* at the ccnlroid of thc tension steel - arc also applicable in the case eccentric compression [mfer CI. 39.1 of the Code]. lo [act, the assumption o distribution of strains [Fig. 13.111 follows directly from the basic assumption t plane secriorrs before berzrlir~gremains plune nfler. bending; this has been valid experimentally. In the case of cccentric compression, howcver, the 'depth' of the (with rcfcrence to the 'highly compressed cdge') can vary from a minimum v x,,,,,,,,(corresponding to e = -) to the maximum value x,, = (i.e., no neutral corresponding to e = 0). The Cock (CI. 39.1) permits %, = 0.0035 to be considered in cases loading cccentricity (i.c., MJP,,) is sufficiently high as to induce some tensi the columo section The li~nitingcondition for this occurs when the result axis coincides with thc edge farthest removcrl from the highly comprcssed cd x,, = D,corresponding to which e = ex = D = eD, as indicated in Fig. 13.1 1.

-

When the loading cccentricity is relatively low, such that the entire sc subjected to (non-uniform) compression and thc NA lies outside thc section (x,, thc Codc (CI. 39,lb) limits the strain in [he higllly cornptessed edge to a between 0.002 and 0.0035 as follows: E,,,

= 0.0035-0.75

,,,,, ,,

&

Torx,, 2 D

where E,, ,,,,,, is the strain in the least conrpressed edge, as shown in Fig. 13.11. I be seen that Eq. 13.18 satisfies. the limiting strain conditions E,,. =. (co~~esponding to &,, ,,,,,, = 0; i.e, x,, = D or e = e), nod E,,, = 0.002 (correspond E,, ,,,,, = 0.002; i.e, x,, = m or e = 0). The point of intersection of these two lit strain profilcs (corresponding to e = 0 and c = e D ) occurs at a distance of 3Dl the 'highly comprcssed edge', and in fact, this point acts like a 'pivot' [or profiles. It serves as a common point thmugl~which all strain profiles (with x. pass, as indicated in Fig. 13.1 1. Using similar triangles, it c m be shown that: e,

[

=0.002 1 +

for.r,, . D

13.5.2 M o d e s of Failure in Eccentric C o m p r e s s i o n Although the term lirnit state of collapse irr comp,rssiorr is generally used by tl (CI. 39) to describe the 'ultimate limit statc' of compression members axially loaded or eccentrically loaded), the actual failure necd not necessarily compression. This is because an eccentrically loadcd colunul section is subjecte an axial comprcssion (P,,)as well as a bending lmorncnt (M,,). The mode of fnjlurc depends on the eccenlricity of loading; i.e., the rela ' magnitudes of P,, and M,,. If the ecccntricity e = M,/P,, is relatively small, the a compression behaviour predominates, and thc co~lscquent failure is term conzpr.ession failure. On the other hand, if the eccentricity is relatively larg flexural behaviour predominates, and the consequent failure is termed tension fa

COMPRESSION

MEMBERS

597

In fact, depending on the exact magnitude of the loading eccentricity e, it is possible to predict whether a 'compression failure' or a 'tension failure' will take place. Balanced Failure

16 between 'compr~sionfailure' and 'tension failure', there cxists a critical failure condition, termed 'balanced failure'. This failure condition refers to that ultimate limit state wherein the yiclding of the outermost row of longitudinal steel on the tension sidc and the attainment of the maximum compressive strain in concrete e,,, = ' 0.0035 at the highly compressed edge of the column occur simultaneously. In other words, both crushing of concrete (in the highly compressed edge) and yielding of steel (in the outermost tension steel) occur simultaneously. 111this context, for design purpose, the 'yield strain' E, is defined simply as that correspouding to the conventional definition of 'yield point' in the design stress-strain curve for steel [refer

+ 0.002

for Fe 250 forFe415lFe500

(13.19)

-

The 'balanccd strain profile' is depicted, along with other strail1 profiles in Fig. 13.11. The corresponding eccentricity in loading is dcnoted e, e,,,=,u,b ; i.c., the eccentricity which results in a 'balanced' neutral axis depth x,, = 4,. b. Evidently, e, < el, < -, where, as explained earlier with reference to Fig. 13.1 1, c~ co~responds to a neutral axis depth x,, = D and e = corresponds to a minin~umneutral axis depth x = x,,,,,,, (when P,, = 0).

-

Compression Failure s than that corresponding to the 'balanced failure' cgndition, i.e., when e < eb, 'yielding' of longitudinal steel in tension does not takc p]%ce,and failure occurs at the ultimate limit state by crushing of concrete at thc highly compressed edge. The compression reinforcement may or may not yield, depending on the grade of steel and its proximity to the highly compressed edge. Tension Failure When the loading eccentricity is greater than that corresponding to the 'balanced f$lure' condition, i.e., when e > eb, failure will be initiated by the yielding of the tepsio~isteel. The outermost longitudinal bars in the tension side of the neutral axis first undergo yielding and sl~ccessiveimler rows (if provided), on the tension side of the neutral axis, may also yield in tension with increasing strain. Eventually, collapse occurs whcn the concrete at the highly compressed edge gets crushed.

-

Load Moment Interaction design strength of an eccentrically loadcd short column dcpcnds on the ntricity of loading. For uniaxial eccentricity, e , the design strength (or resistance) two components: an axial compression component, P,,,, and a corresponding axial moment component, M,,R= P,,Re.

598 REINF O RC ED CONCRETE DESIGN

DESIGN OF

COMPRESSION

MEMBERS 599

As seen in Section 13.5.1, there exists a unique stwin profile (and n location) at the ultimate limit state, co~respondingto a given eccentricity [Fig. 13.111. Corresponding to this dislribution of strains ('strain compatibili stresses in concrete and steel, and hence, their respective resultant forcest C, can be determined. Applying the condition of static equilibrium, it follows, two design strength components are casily obtainable as:

PUR= Cc + Cs and

M,,R= M,+ M,

where M, and M, denote the resultant moments due to C, and C, respective respect to the centroidal axis (principal axis under consideration). From the nature of the equilibrium equations [Eq. 13.20, 13.21], it observed that, for a given location of the neutral axis (n,,lD),the design values P,,, and M,,, can be directly determined, and the eccentricity e resulting in such a NA location can be deduced. However, given an arbitrary v e, it is possible to arrive at the design strength (P,,Ror M,,, = P,,n e ) using Eq. ollly after first locating the neutral axis - which can be achieved by c tnolnents of forces C, and C, about the eccenttic line of action of P,,,, static cquilibriun~. Unforlunalely, the expressions for C, and C, in (derived in Section 13.5.4) are such that, in general, it will not be possib closed-form solution for x,, in terms of e . The relatio~lshipis highly requiring a hial-and-error solution. DESIGN INTERACTION CURVE

Interaction Curve

The 'interaction curve' is a complete graphical representation of the design s of a uniaxially eccentrically loaded c o l u ~ ~of m given proportions. Each point curve corresponds to the design stmngth valucs of P , and Mu,associated specific eccentricily (e) of loading. That is to say, if load P is applied column with an eccentricity e, and if this load is gradually increased till the tilt limit state (defined by the Code) is reached, and that ullimate load at failure is by P,, = P,,, and the corresponding molnellt by M,, = MilR= PUR e , then the coord (M,,,, P,,,)' fonn a unique point on the interaction diagram (such as point Fig. 13.12). The interaction curve defines the different (M,,R, P,,n) combinatiol all possible eccentricities of loading 0 5 e < -. For design purposes, the calcu of M,,, and P,,R are based on the design stress-strain curves (including the safety factors), and the resulting interaction curve is sometimes referred to de.sign inreruction curve (which is different from the chamcteristic interaction Using the design interaction curve for a given colulm~section, it is pas makc a quick judgement as to whether or not the section is 'safe' under

Some of the longitudinal steel may be subjected to tension, rather than compressiol term C, helx denotes the net force (assumed positive if compressive) considering all tllc b the section. ? It is customary to use the x - axis far M,, values and they - axis for P,, values.

(P" = P"", M" = MUR1

e<

Fig. 13.12 Typical P,-

eb 3

'compression failure'

Muinteraction diagram

DESIGN OF COMPRESSION MEMBERS

600 REINFORCED CONCRETE DESIGN

axis is located outside the scction (x,, > D), with 0.002 < E,,, < 0.0035. Fo the NA is located within the seetiott (z,.< D) a 1 ~ 1E,,. = 0.0035 at th compressed cdge' [Fig. 13.1 I]. Point 2 represents a general casc, with. axis outside the section (e < c~ ). The point 4 in Fig. 13.12 corresponds to the baloncerl fai1rrr.e condi e, and x,,= x,,, [refer fig. 13.111. The design strength values for t failure' condition arc denoted as Pllband M,,,,. FOKPilR< Pllb(i.e.: e . mode of failure is called fer~sion failure, as explained earlier. It may b M.b is close to the maximum' value of nltimate moment of resistane given scction is capable of, and this value is higher than the ultimat resisting capacity M , , under 'pure' flexure conditions [point 5 in Fig, 13. The point 5 in fig. 13.12 corresponds to a 'pure' bending condition P,,R=0); the resulting ultimate moment of resistance is denoted M,, corresponding NA dcpth takes on a minimum values,,, ,,,,,,.

13.5.4 Analysis for Design Strength In this scction, the detailed calculations for determining the desigt s f m g f h of a uniaxially eccentrically loaded column with a rectangular cross-section (6 x D)is described in detail. The ootatioo D denotes the 'depth' or the rectnngular section in, the planc of bending, LC., either D,or D,, depending on whethcr respect to the major axis or minor axis, and the notation b d (width) of thesection (in the perpendicular direction), The basic procedure cross-sectional shapes (including eirculx sections) is sinular, nod this is dem in Example 13.8 for an H-shaped section. This procedure can also be extendedt lxge tubular towers (such as chimneys), albeit with some tnodificalions [Ref. 13,.1&] As explained in Seetioti 13.5.3, the design sbength of an column is not a unique value, but comprises infinite sets of va (cotresponding to 0 < e < -) - all of which are dcscribable by mcans of a si curve, termed the design iniervcfion curve [Fig. 13.121. It was the analysis for design strength basically entails two conditions: strain compat [Fig. 13.111 andequilibrium [Eq. 13.20, 13.211. 'l'hc dirtr~bul~o~l of itl:m.; I I I tlla rcctnngul~rcolumn ~ : t ~ o.lnJ n the concspunding ;, .' (.c ~ l . m ~ l r s s l v e.trc~sr.\ ) t n m1:letr .ire dclwxcd i n Fig. " 13.13 Two cliffercnt c a w. 0 (compressive) [refer Eq.13291. However. it is sufficientlyaccurate to consider fci = 20 MPa or 25 MPa for this purpose and thereby n~ðe curvcs applicable for all grades of concrete [Ref. 13.12].

622 REINFORCED CONCRETE DESIGN

DESIGN OF

.

unsymmetrically arranged reinforcement in rectangular sections; non-rectangular and non-circular sections - such as Lshaped, T-sha shaped, cross shaped sections, etc. In such cases, it becomes necessary to construct proper interaction dia order to obtain accurate and reliable solutions.

COMPRESSION

MEMBERS 623

design chart used refers to the case of "equal reinforcement on four sides"

-

area required = 3960 (1232 x 2) = 1496 m2 Ide 4 22 $in two inner rows: area = 380 x 4 = 1520 mm2 > 1496 mm2

-

factored load of 1400 kN and a factored moment of 280 kNm with resp major axis. Assume M 20 concrete and Fe 415 steel. SOLUTION

ng 8mm ties, effective cover = 40+8+(28/2)= 62mm = 60-OK ailing is shown in Fig. 13.22. Details of transverse reinforcement are also

8 4 ties O 200 clc

ble modifications to the reinforcement provided.

.

.

Fig. 13.22 Example 13.12

Given: b = 300 mm, D = 600 mm,Ak= 20 MPa,f, = 415 MPa, P. = 1400 ldrl, M,, = 280 kNm Arrangement of bars: as D = 600 mm, the spacing between tie corner bar exceed 300 mm; hence inner rows of bars have to be provided to satisfy det requirements [refer Section 13.3.31, Assuming two or more inner row SP: 16 Chalis for "equal reinforcement on four sides" can he made use [Fig. 13.21(b)]. Assuming an effective cover d' = 60 mm. =, d'/D = 601600 = 0.1

pu =- P" f,bD

.

=

ctive cover d'= 40 t 8 t 14 = 62 m m arrangement of bars in this case conforms to "reinforcement disuibuted

lf,k=2.213/20=0.1106

-

0.11

1400x10' = 0,3p 20 x 300 x 600

Mar 280 x lo6 - 0.130 m,, = -f , b ~ ' - 2 0 x 3 0 0 ~ 6 0 0-~ Referring to Chart 44 (&ID= 0.10) of SP : 16, it can be observed that, coordinatesp,, = 0.389, m,, = 0.130 would lie on a design interaction curve with pKk=O.ll s p , e q d = o . l lx 2 0 = 2 . 2 = , A , ,ryd = 2.2 x 300 x 6001100 = 3960 mm2

Referring to Chart. 34 (d'/D= 0.2) of SP: 16, it can be seen that the point

P,,= 0.389, mu = 0.185 lies outside the design interaction curve envelope forplf, = 0.11 and d'/D= 0.09. The value of pKk corresponding to p , , ~= 0.389 and nz,,~ = 0.185 is given by: @lf,kLeqd = 0.18 > @Kk)p,ovi~ed = 0.11. Hence, the given section is unsafe. Corresponding to (plf,),,,

P , ~= , ~0.175 x 20 = 3.5

= 0,175,

DESIGN

626 REINFORCED CONCRETE DESIGN

OF

COMPRESSION MEMBERS 627

Alternatively, the resultant eccentricity e = MJP,, may be obtained [refer Fig. 13.24(b)] as: (13.37) When the column section (including the reinforcement) is axisymmetric (with reference to the longitudinal axis) - as in a circular column - the resultant axis of bending is also a principal axis [Fig. 13.%(c)]. In such a situation, the case of biaxial bending simplifies into a case of uniaxial bending. The neutral axis, in this instance, will remain parallel to the resultant axis of bending.

Fig. 13.25 Analysis of design strength for a given location of neutral axis !

.I...

.'

,.'

I

,,:, io ,' !

.'

of bending

e

i er

,.......... .. e7

Fig. 13.24 Resultant eccentriclty of loading However, in the more general case of non-axisymmetric reinforced concrete column sections, the neutral axis is generally not parallel to the resultant axis of bending [Fig. 13,24(d)]. In fact, the determination of the exact neutral axis location is a laborious process of trial and error. For a given neutral axis location, however, the failure strain distribution can be drawn (with the same assumptions as in the case of uniaxially eccentric compression) [Fig. 13.251.

13.6.2 interaction Surface f o r a Biaxialiy Loaded Column Various simplified procedures for the desigu of biaxially loaded colunms have been proposed [Ref. 13.14, 13.151 and adapted by different design codes. Most of thcse simplified procedures are bascd on an approximation of the ir~reruction surface, which may be visualised in a three-dimensional plot of P.# - M,, - MjCy [Fig. 13.261. The surface is genclsted as the envelope of a number of design interaction curves for different axes of bending. Each point on the interaction surface [Fig. 13.261 corresponds to values of P,,,, M,, and M , , obtained from the analysis of a chosen neutral axis location and orientation, such as the one in Fig. 13.25. The design interaction surface can be conddn-ed to be a failure sqfoce in that the region bounded within this surface is a 'safe' region and any point (P,,, M,,. M,,) that lies outside the surfacc is 'unsafe". The traces of the interaction surface on the x-z and y-z (vertical) planes correspond to the design interaction curves for uniaxial eccentricity with respect to thc major and minor principal axcs respectivcly. In order to avoid confusion, the notations llsed for the design flexural s t ~ n g t h under uniaxial eccentricity and undcr biaxial eccentricities, the lollowing norations shall be uscd in the context of biaxial loading of colulluls: M,,R. design flexural stccngth with respect to ~najoraxis undcr binsiul loading MCd, design flexural strength witli respect to minor axis under binxial loading

--

'That is, the comsponrling probability of failure is unaccepmble, according to the Codr.

DESIGN

OF

COMPRESSION MEMBERS 629

It is interesting to note (in Fig. 13.26) that the trace of the interaclion surface on a horizontal plane (parallel to the x-y plane) at any load level P,, is also an interaction curve - depicting the interaction between the hiaxial bending capacities M,,R,rand Mid).. Such an intcmction curve is sometimes referred to as a load contour, as all the points on the curve pertain to a constant axial load level.

13.6.3 C o d e Procedure f o r Design of Elaxially Loaded C o l u m n s The simplified method adopted by the Code (C1.39.6) is based o n Bresler's formulation [Ref. 13.141 for the 'load contour' - whereby an approximate elationship between M,,R,rand MuR,Y (for a specified P,, = P,,R) is established. This elationship is conveniently expressed in a non-dimensional form as follows: (13.38) where M,,r and M,, denote the factored hiaxial moments acting on the colorno, and (as explained earlier) M,,, and M,,y~ denote the uniaxial moment capacities with reference to the major and minor axes respectively, all under an accompanying axial load P,, = P,,R. It may be noted that M,,, M,, (and P,) are measures of the load effects due to exteiml loading on the structure, whereas M,,r,, M,,yr(and P,,,) are measures of the inherent ,rsistar!ce of the column section. a,, in Eq. 13.38 ia a constant which depends on the factored axial compression P,, and which defines the shape of the 'load,contour' [refer Fig. 13.271. For low axial load levels, the load contour (in non-dimensional coordinates) is approximated as a straight line; accordingly a,, = 1. For high axial load levels, the load contour is approximated as t h e quadrant of a circle; accordingly a,, = 2. For moderate load levels, a, takes a value between 1 and 2, as shown in Fig. 13.27(a). In order to quantitatively relate a,, with P,,, it is condnient to normalise P,, with the maximum axial load capacity of the column (under 'pure compl-ession'). This was denoted as P, in Section 13.4.3, and defined by Eq. 13.16, with slightly different expressions for different grades of steel. In the context of biaxial loading, the Code (Cl. 39.6) uses the notation P,,, (instead of P,,,), and suggests the following rounded-off version of Eq. 13.16, applicable for all grades of steel: Mu,= P,.C'

P,,, = 0.45f,xA, -P,,,=

Fig. 13.26 Interaction surface for a biaxially loaded column M,,, =design flexural strengtll with respect to major axis ondcr ti~dn.~ial loadir~g

(i.e., e, = 0) M,,?, -design flcxural strength with respect to minor axis under rr,~iaxialloading (i.e., e, = 0) Thc notations and their respective meanings are depicted in Fig. 13.26, corresponding to an axial compression P,,= PZzR.

+ 0.75f,A,,

0.45f,kA, +(0.75&-0.45fc3A,,

(13.39)

where A, denotes the gross area of the scction and A, the total area of steel in the section. a,, = 1 for P,,IP,,, < 0.2; a,, = 2 for P,,/PP > 0.8; and a,, is assumed to vary linearly for values of P,,IP,,, between 0.2 and 0.8 as show? in Fig. 13.27(b). Accordingly,

I'"

a,,= 2.0

for p,,/ 1.0 Hence, the glven load is found to marginally exceed the safe limit prescribed by tlte Corlc (by 8%).

(b)

Pig. 13.29 Behaviour of slender columns

13.7 DESIGN OF SLENDER COLUMNS 13.7.1 Behavlour of Slender Columns , , ~8''

::

As discussed in Section 13.1.3, compression members am categorised as being either short or slender (long), depending on whether slende~nesseffects can be ignored or need special coasideratio~t. It is also explained in Section 13.1.3 that the slenderness mtior (I,lD,, /,ID,) provide a simple basis for deciding whether a column is short or 'slender'. The behaviour and design of short columns under axial, uniaxial eccentric and biaxial eccentric loading conditions have been extensively described in Sections 13.4, 13.5 and 13.6 respectively. This section describes the behaviour of slender columns, and shows how this beltaviour increasingly deviates front the short column behaviour with increasing slenderness ratios. To begin with, a simple example of a pin-ended column with an eccentrically applied load [Fig. 13.29(a)] is considered. The height 1 between the pinned ends is the 'unsupported length', which, in this case, is also equal to the 'effective length' [refer Fig. 13.3(a)]. By considering different heights of tlte colullm, with the same cross-section, the effects of different slenderness ratios can be studied. Subjecting the column to a gradually increasing load P , applied at an eccentricity e (with the undeflected longitudinal axis), the behaviour of the colunm call be observed until failure. Due to the applied eccentricity e, 'primary moments' M , = Pe are developed not only at the end sections of the column, but all along the height [Fig. 13.29(b)]. The bending of the column causes it to deflect laterally, thereby introducing additional displacement (load) dependent eccentricities. If the lateral deflection of the longitudinal axis is denoted as A, then the total eccentricity is e + A, and the total 8 moment M at any section is given by M=P(e+A) (13.42a)

(c)

:

,

.

where PA is the 'secondary moine~~t'(also .called 'P-A mnon~ettt'), which has a Variation along the height of the column that is identical to that of A [refer Fig. 13.29(a), (b)]. The maximu~nvalue of A (i.e., A,,,,), and hence the maximum "slue of the total moment M,,,, = P(e + A,,,,) occurs at the illid-height section of the colum~~. It should be noted that the laleral deflection A,,,, is not only due to tlle curvature produced by the primary momcnt M,,,, but also due to the P-A moment. Hencc, the variation of M,,,,witit P is nonlinear, with M,,,,, increasing at a faster ratc as P increases. The axial thrust P eficctively reduces the flexural stiffness of the column ('beam column'), and, in the case of a very slender column, it may so happen that the flexural stiffness is effectively reduced to zero, resultillg in an instability (buckling) failure, On @e other hand, in the casc of a very short colulm~,the flexural stiffness is $0 high that the lateral deflection A is negligibly small' ; consequently, the P-A moment is negligible, and the primary moment My, alone is of significaltce in such a case. Fig. 13.29(c) shows the axial load-moment interaction curve (at the ultimate lilujt state) for the colunnt secrion. This curve, therefore, represents the strength of the column. Also shown in Fig. 13.29(c) axe three different loading paths OA, OB, OC that are possible (for different slenderness ratios) as the column in Fig. 13.29(a) is loaded to failure, with incmasing P (and hcnce, M) and constant eccentricity e. In the case of a very short column, A,,,., = 0 (as explained earlier) and M,,,, = Pe. The resulting P - M path is linear, as indicated by the line OA in Fig. 13.29(c). The 'Theoretically, A,,,, = 0 only if the effective length 1, = 0 or if e = 0 (pure axial Loading). In practical 'short' columns, some lateral deflection is unavoidable, particularly at high eccentricities of loading. However, it is expected that the P-A moment in a short calulnn will not exceea about 5 percent of the p r i m x y mnoment, and so may be neglected.

DESIGN termination of this line at thc point of intersection A with the interaction (failure) curve indicates the failure of the column at a load, say P = P, and a moment M , , , = P,e. The failure occurs by thc cmshing of concrctc at thc section of maximum moment. Had the column been longer (and hence, 'slender'), with increasing load P, the deflection A,,!, is no longer negligible, and the momcnt M,,,, = P(e + A,.,) will vary nonlinearly with P, as indicated by the line OB in Fig. 13,29(c). Failure occurs at a load P = P Iand a moment M,,, = Pl(e +A,); this is represented by the point B on the interaction curve. In this case. Ple and P,A, denote respectively the primary moment and secondary (P-A) momcnt at failure. As shown in the figure, the secondary moment can become comparable to the primary momcnt ill magnitude at the ultimate limit. state. Fhrthermore, comparing the loading paths OA with OB, it follows that although the column section and thc eccentricity in loading arc identical in the two cases, the merc fact that one column is longer than the other can result in a reduction in the load-carrying capacity (as well as the primary inomcnt resistance). In botli cases, the final [ailwe will be a material failure -either a 'compression lailurc' or n 'tension failure' depending on which parts of the interaction curve the points A and B lic [rerer Section 13.5.21. Most columns in practical building frames are cxpected to have this type of failure at the ultimate limit state. If the column in Fig. 13.29 is very long, the increase in lateral deflection A,,>,,,may be so excessive that the load-moment path corresponds to OC, with (IPIdM reaching zero at the point C. In this case, the column is so slcndcr that it fails by instability (buckling) at a relatively low axial load P,. This type of failure may occur in very slender columns in mibraced frames.

OF

COMPRESSION MEMBERS 637

sufficiently high (curve '2' in Fig. 13.30). the total moment to be considered in design (i.e., including the additional moment PAz) may exceed M2 This is less likely in columns bent in double curvature [Fig. 13.30(b)]. In fact, the chances of a given slendcmess resulting in a peak design moment larger than M, fall off significantly as the ratio MIIM, drops below about +0.5 and approaches the limit of -1.0. Thc possible amplification in bending moment (over the primary moment M2) on account of lateral displacements (relative to the chord joining the colunm ends) is tenned as member srnbilify effeir.

-

Braced Slender Columns: Member Stability Effect As explained in Section 13.2.3, a 'braced column' is one $iich is not subject to sidesway, i.e., thcre is no significant relative lateral displacement between the top and bottom ends of the column. The pin-jointed column of Fig. 13.29 is a simple example of a braced column. In eeneral. the ends of a braced column (which forms Dart of a 'braced frame') are partially restrained against rotation (by the connecting b The primary moments MI and M2 that are applied at the two cnds of the colut determined from a 'first-order' sfructural analysis; i.e.. analysis which assume elastic behaviour, and neglects the influence of change in geometry of the fra to deflections. The colunm may be bent in single curvarrrre or double cur depending on the directions of MI and M2 [Fig. 13.301. The notations MI an generally refer to the smaller and larger column end moments, and the ratio considered positive if thc column is bent in single curvature, and uegative if in double curvature. If M,IM, = +1.0, the c o l o m ~is bent in symmetrical single cu slcndeniess in the c o l u m ~will invariably result in an increased moment. Howcver the more keneral case of unequal end moments (MIIM, ;e LO), it is not nece slenderness will result in a peak moment in the column that is grcater than the primary end moment Mi - as indicated by the curves labclled "I" in Fig. 13.3 however, the column is very slender, and the consequent lateral

-

(a) single curvature

(b) double curvature

Flg. 13.30 Braced columns: member stability effect Thus, the criticality of slenderness effects is also dependent on the ratio MIIM2. The ACI Code [Ref. 13.11 recommends that slenderness effects may be ignored (i.e., the column may be designed as a 'short column') if, for a braced column, 1,Ir < 34 - 12 Ml/Mz

(13.43)

where 1, is the effective length and r the radius of gyration. Thus, the sle?derness ratio (ldr)limit for short columns lies in the range 22-34 in single curvature and 3446 in double curvature. It is shown [Ref. 13.161 that this slenderness limit [Eq. 13.431 corresponds to effective lengths for which the ultitn?te axial load capacity, including 'member stability' effect, is at least 95 percent of the axial compressive strength of the cross-section. Unbraced Slender Column: Lateral Drlft Effect As explained in Section 13.2.3, an 'unbraced column' is one which is subject to sideway (or 'lateral drift'), i.e., there is significant lateral displacement between the

638 REINFORCED

C O N C RETE DESIGN

DESIGN OF

"p and of the colutm. The lateral drift may occur due to the ac lateral loads, or due to gravity loads when the loading or the frame is asytmnetri

COMPRESSION

MEMBERS

639

d columns, the moments at the colutnn ends aremaximum, and these are due primary monmits enhanced by the lateral drift effect alone. motllcnt amplification possible due to lateral drift effect in an unbraced urn is generally much morc than that due to. mne~nberstability effect in a braced Ft~rther,as cxplaitad in Section 13.2.3, the effective length of an unbraced is tnuch more than that of a braced column with the same utlsupported letlgtll. columns in unbraccd lrames are wcaker than similar columns in hraced

7.2 Second- Order Structural Analysis of S l e n d e r c o l u m n S t r u c t u r e s The tnaitl problem with slender colunui design lies in deter~llinin~ the factored moments (including P-A effects) to be considered in design. In other words, the p b l e m is essentially one of structural analysis, rather than structural dcrisrt. The principles of dcsigtling a column section under a givcn factored axial coinpressioll P,, [descrihcd in Section 13.61 remain the same for both and factored molnetlts M,,.?, sllort columns and slcnder columns; the only difference is that M,,, and M,,, must include secondary moment components it1 slender colutm~design, whereas these secotldary moment components (bcing negligible) al-e ignored in short column design.

(a) sway frame

Rigorous A n a l y s i s

(b) swayed

column

(C)

In general, the Code (Cl. 39.7) broadly recomnletlds that when slehder columns are involved in a winforced concrete structure, a detailed 'second-order' structural analysis should bc carried out to determine the bending moments and axial forces for which the slender colomns are to be designed. Indeed, such a rigorolls analysis is particularly desirable for slender columns in utlbraccd frames. Such analysis lllust take itlto account all slenderness cffects, viz. the influence of column and frame deflections on mnoments, cffects of axial loads and efkcts of sustained loads. Realistic ~iloment-curvautrerelationsl~ipsshould be made use of. The dctails of procedures for second-order analysis lie outside the scope of this book; these details are presented in Ref. 13.17- 13.19. It should be noted that the prbtciple of su~xvposilio~~ is not valid in second-order analysis, and for this reason, the load effects due to different load combinations cannot be obtained by an algebraic sumning up (with nppmpriate load factors); each load co~nbitlation should be investigated separately. This requires substantial computational effort.

forces

13.31 Unbraced columns: lateral drift effect

Considering the simple portal frame of Q. 13.31(~) (in ,which the is assumed to be infinitely rigid, for convenience), the lateral drift (or sideway) of colutlln is the relative translational displacetnent A (= A, + A,) between the of the column. The additional moments at the column ends caused by actiol, of the vertical load acting on the deflected configuration of the unbraced is termed Lateraldrift effect. In unhraced columns, the action of pritnary moments (M,, M2) getlerally results in 'double curvature', which is further ellhanced by the lateral effect. In addition, there is the 'member effectv(described on account of the lateral displacements at ~ ~ o i nalong t s the lengtll of columl relative the chord joining the column ends [Fig. 13,31(d)]. H ~ generally, ~ for ~

13.7.3 C o d e P r o c e d u r e s f o r Design of S l e n d e r C o l ~ m n s

~

111routille design practice, only first-order structural analysis (bascd on the linear elastic thcory and undeflccted fuame geometry) is perfor~uetl,as second-order a~lalysis is colnputationally difficult and laborious. In recognition of this. the Code recommends highly simplilied lprocedures lor the design of slender columns, which either attempt to prcdict thc increasc in mornents (over primary ~llolllents),or. \n the ,reduction 111 strength, due to slenderness effects. ~equivalently, ~

640 REINFORCED CONCHETE

DESIGN OF COMPRESSION MEMBERS 641

DESIGN

S t r e n g t h R e d u c t i o n Coefficient M e t h o d This is a highly simplified procedure, which is givcn in the Code for the working stress mcthod of design [refer Section 13.4.31. According to this procedure (B-3.3 of the Code) the permissible srresseu in concretc and stcel [Eq. 13.15, 13.161 are reduccd by multiplication with a strength rerl~rc~ior, coefficierif C,, given by:

where d is thc least lateral dimension of the colulm~(or diameter of the corc in a spiral column). Alternatively, for more exact calcrilations,

The essence of this method lies in a simple formulation for the determinalion of the additional eccentricities e,, .e, In the basic formulation, the P- A effect in a braced slender col~lmnwith pin-joined ends [Fig. 13.29aj is considered. The 'additional eccentricity' e, is equal to A,,,, in Fig. 13.29(a), which is a function of the curvatures to which the column is subjected. If the maximum curvature (at mid-height) is denoted as rp,,,, it can be shown [refer Fig. 13.321 that A,,,, lies between q7,,,l2112 and rp~,,,,,l2l8,the former limit corresponding to a linearly varying curvature (with zem at thc pin joints and a maximum of yl,, at midheight) and the latter corresponding to a constant curvature along the column height [Fig. 13.32(b),(c)j. Taking an average value,

,--q~ = MIEI

where r,,,i, is the least radius of gyration of the column. There is some ambiguity in Eq. 13.44(a), (b) regarding the plane in which ihc eflective lengtll 1, is to be estimated. This can be resolved by considering (/.Id),,,, in Eq. 13.44(a) and (l,lr),,,, in Eq. 13.44(b) ic., considering the n~aximumeffective slenderness ratio of thc column. It is recommended in the Explanatory Handbook to the Code [Ref. 13.71 that instead of applying thc strength reduction lactor C, lo the 'permissible stresses', this factor may bc directly applicd to the load-carrying capacity estimated fox a .

.

the case of axial loading (without primary bending mo~nents).This is demonstrated in Example 13.17. Additional M o m e n t M e t h o d The method pl-escribcd by the Code (CI. 39.7.1) for slender c o l u m ~design by the limit state method is the 'additional moment mcthod" , which is based on Ref. 13.20, 13.21. According to this method, every slender colutm~should be designed for biaxial ecccntricilies which include the P-A moment ("additional moment") , components e,, a MJP,, and e , = MJP,, :

LL M

DEFLECTION

CURVATURE

(4

(b) case 1

(c) c a s e 2

Fig. 13.32 Relation between A,,, and rpIrarIn a pin-joined braced slender column

-

-

Here, M , , and M,, denote the total design momcnts; M,,,, M , ,denote the primaq factored motnenlsd (obtained fiom first-order structural analyses); and M,, M, dcnote the ndrlitior~al,nornerrts with reference to bcnding about the major and minor axes respectively.

'

An alternative method called the 'moment magnification method' is adopted by the ACI and Canadian codes The primary tonmnts should not be less than tlme corresponding to the miniinum eccentricities specified by the Code.

Failure of the column at the ultimate limit state is expected to occur at the section corresponding to p,,,. By making suitable assumptions, rp,, can be expressed in terms of the failure strains E , , and E,, in concrete (at the highly comnpressed edge) and steel (in the outermost row) respectively, as shown in Fig. 13.33. The values of e,,, and E,, evidently depend .on the factored axial load P,, (as cxplaincd in Scction 13.5.1): this determines the location of the point of failure, marked B in the interaction curve in Fig. 13.29(c).

642 R EIN F O R C E D CONCRETE

DESIGN

P"

Fig. 13.33 Determination of curvature from failure strain profile

..

'Assulllillg that E l , = 0.0035 and E, = 0.002', d'= 0 . l D and further assun,ing (ratha conservatively) that the additio~~al moment comprises about 80 percent of the tota nlonlent, 'PnIax

2

0.0035 +0.002 x 0.8 = 1 0.9D 2000

-

Conlbiniw Eq. 13.47 with Eq. 13.46, the following expression for the additiollal cccentricity ratio eJD is obtained:

--

e,lD = O/D) 2000 Accordingly, the following expressions for additional ~ n o ~ n e ~M,, lts Eq. 13.45a, bl are obtained, as given in the Code (CI. 39.7.1): M-. = p,, em = M, = P,, e,

M,

~e effective length I, in Eq. 13.49 to extend the application of the formulation to the various boundary conditions (other than the pinned-end condition) that occur in practical columns including unbraced columns. It is reported [Ref. 13.71 that the use of Eq. 13.49 has been validated with reference to a luge number of experimental tests [Ref. 13.201. It is seen from Eq. 13.48 and Eq. 13.49 that the e.lD ratio increases with the square of the slenderness ratio IJD; edD has a minimum value of 0.072 for IJD = I 2 (transition between 'short column' and 'slender column') and a maximum Value of 0.450 for 1JD = 30 (recommended limit for unbraced columns) and 1.800 for 1;li) = 60 (braced column). :' It should be noted that Eq. 13.49 relates to the 'additional moments' to b e considered in addition to the maximum factored primary moments M,,, M,, in a column. Under eccentric loading, these primary moments should not be less than those corresponding to the minimum eccentricities specified by the Code. Where a brimary moment is not considered, i.e., taken as zero, (as under axial loading), it should be ensured that the corresponding additional moment is not less than that computed from considerations of hinimum eccentricity. The derivation of Eq. 13.49 issumes that the column is braced and bent symmetrically in single curvature; some modification is required when the primary moments applied at the column ends are unequal andlor of different signs. Further, it is assumed that the axial load level corresponds approximately to the 'balanced failure' condition P,, = P,& Eq. 13.49 needs to be modified for other axial load levels. Hence, the Code recommends the following modifications to b e incorporated with the use of Eq. 13.49 (and Eq. 13.45) for the design of slender columns in general: For P,, > the failurc mode is one of 'compression failure', and the corresponding elD ratio is low. At relatively high axial loads, the entire section may be under compl.ession, suggesting low curvatures. Hence, the use Of Eq. 13.49 in such situations can result in highly conservative results. The additional moments M,, M,,ygiven by Eq. 13.49 may be reduced by multiplying factors (refer C1.39.7.1.l of the Code) defined as:

[in

(le~,)'

c,

D,.( W Y ) 2 = ---2000

where and levdenote the effecrivc lenglhu, and D, and D, denote the depths rectangular colulnn section with respect to bending about the major axis and axis respectively. It may be noted that the height I in Eq. 13.48 has bcell replace

' This aPPl.oximatelycorrespo~~dingto the 'balanced failure' condition, wilexby&, = F, tg cracked section. For deflection calculations, the mean steel strain should be collsidered including rhe effect of 'tension stithing' (refer chapter 10).

where P,,is the maximum 'pure compression' swength of the column and Pub,, and P,,b,? correspond to the axial strength corresponding to balanced failure with respect to bending about the major axis and minor axis respectively. P,, is readily obtainable from Eq.13.39 and Pubfrom the interaction curve (refer Fig. 13.20) corresponding to a design tensile stress off,d = 0.87 f, in the outermost layer of steel. It can be seen that k varics lincarly from zero (for P,, = P,,,) to unity (for P. = P,d and is a highly simplified fornlula. It should also be noted that Eq. 13.50 is not applicable for P,, < P,,& i.e.. k, = 1 for P,, < P,,b.

For braced colualns subject to unequal primary moments MI, Mz at the two ends [Fig. 13.30(a)1, the value of M,, to be considercd in the computation of the total momentk,, in Eq. 13.45 may be taken as:

where Mz is the higher column cnd moment. As mentioned earlier, with referenc; to Fig. 13.30, MI and MZ are considcred to be of opposite signs.if the column is bent in double curvature. In the case of braced columns subject to double curvature, it is possible that the use of Eq. 13.51 in Eq. 13.45 may result in a total moment k , , that is less than Mz; this obviously, cannot be allowed. Hence, a further condition needs to be imposed:

k,, 2 Mz for braced columns (13.52) In the case of unbraced colurnns, the lateral drifl cffect (hitherto not considered) needs to be included [Fig. 13.311. An approximate way of accounting for this is by assuming that the additional moment M, (given by Eq. 13.49') acts at the column end whcre the maximum primary tnometlt M2 is operational. Hence, for design purposes, the total moment k , , may be taken as:

fi,,= M~+ M,

for uubraced columns

(13.53)

,

e,o

-

(pJ,,,u = Cr P,, F0 . 8 3 7 ~ 2290

= 1917 kN safety of the column under this factored load, combined with minimum may now be verified by the additional moment method given in the Code for LSM.

.

Additional moment method i .hfinlmurneccentricities + ? L 7000 = + -500 =30,67mm>20mm e,,,,,= 500 30 500 30 300 1 D~ -+ - = 24.00 tnm > 20 mm e,,,,,,= + 500 30 500 30 Primary nlomenfs loading, M,, = M , , = 0. However, it %nustbe ensured the column is under that the totalmolnents M,, , M , , should not be less than those due to

-

Determine the maximum factored axial load-carrying capacity of the column in Fig. 13.14(a), given that the column is 'braced' against sideway, and has an unsupported lengtll of 7.0 m. Assume effective length ratios k, = k, = 0.85. SOLUTION

- 7000

.

EXAMPLE 13.17 -

considering short column behaviour (with dimensions satisfying CI. 39.3 of the Code), = 0.4fcbA, + (0.67f, - 0.4fck)As = 2290 kpl (as determined earlier in Example 13.9). Considering slender column behaviour,

.

- -

corresponding minimum eccentricities.

Additional nlornents Without modification factors: = D~((~JD~)~/~OOO = 500 (1 1.90)?2000 = 35.40 e,, = D, ( i e ~ ~ y ) 2 ~ = 2 0300 0 0 (19.83)212000 = 58.98

Given: (refer Example 13.5): D, = 500 mm, D, = 300 mnl, A, = 2946 m z ,

frx = 25 MPa,..,f.= 415 MPa.. ~~-

Also, I = 7000 mm, k, = k, = 0.85. Slenderness ratios: 1, = 0.85 x 7000 = 5950 nun L,JD, = 59501500 = 11.90 < 12 I, = 0.85 x 7000 = 5950 mm 3 l,ID, = 59501300 = 19.83 > 12 Hence, the column has to be treated as a slender. column. Strength reduction coefficient method Extending the sll.ength reduction coeficienr methorl given in the Code (B-3.3) for WSM to LSM,

-

where P,,, = 0.45fCkA, + (0.75fy- 0.45fcx)As = (0.45 x 25 x 300 x 500) + (0.75 x 415 - 0.45 x 25) X 2946 = (1687.5 x l o 3 + 883.8 X lo3)N= 2571 kpl p = . a,"," 445 kpl.. p ,."., = 4 7 0 (as~ determined inExamples 13.9, 13.11) =, k.. = (2571 - 1935)1(2571- 445) = 0.299 = (2571 - 1935)/(2571 - 470) = 0.303 M,,y= p,, (kme,) = 1935 x (0.299 x 0.0354) = 20.5 M m M, = P,, (k,e,,) = 1935 x (0.303 X 0.05898) = 34.6

.

4

mm

." 1.25 - 19.83148 = 0.837

I t is inadvisable to apply the reduction factor k (given by Eq. 13.50) for onbraced columns

DESIGN OF COMPRESSION MEMBERS

649

P,JP,..= 150013087 = 0.486 (which lies between 0.2 and 0.8)

Fig. 13.34 Example 13.18 Check additional moments '

Assuming a clear cover of 40 mm, d' = 40 + 8 + 2812 = 62 d'/Dx = 0.155 = 0.15 and d'/D, = 0.207 = 0.20

-

Referring to Charts 45 (d'/D= 0.15) and 46 (d'/D= 0.20) of S P :16, the loads Pub,.,P.b,> at balanced failure can be determined by considering the stress levelf,,= 0.87& (marked on the interaction curves). Corresponding top& = o . I x , for d'/D, = 0.15, P,,&kbD = 0.07 =, P , , ~=, ~252 k~ for d'/D,. = 0.20. P,,&,/f,xbD= 0.03 P,,~,,= 108 l c ~

-

P I , = 0.45fh A, + (0.75& - 0.45fJA, = (0.45 X 30 X 300 x 400) + (0.75 x 415 - 0.45 x 30) x 4928 = (1620 X l o 3 + 1467 x lo3) = 3087 kN Mod~cationf(zctors: k".? = "Y

CZz- &,=78.1 M m

REVIEW QUESTIONS What is meant by slenderness ratio of a compression member and what are its implications? 13.2 Distinguish between (i) unsupported length and effective length of a compression member; (ii) braced column and unbraced column. 13.3 Why does the Code require all columns to be able to resist a n~inirnurrr eccentricity of loading? 13.4 Why does the Code specify limits to the minimum and maximum reinforcement in columns? 13.5 A short column, 600 nun x 600 mm in section, is subject to a factored axial load of 1500kN. Determine the minin~unrarea of longitudinal steel to be provided, assuming M 20 concrete and Fe 415 steel. 13.6 Enumerate the functions of the vansverse reinforcemen& in a reinforced concrete column. 13.7 Explain the limitations of the traditional working stress method with regard to the design of axially loaded reinforced concrete column. 13.8 Compare the behaviour of tied columns with spiral columns, subject to axial loading. 13.9 Sketch a typical axial load - moment interaction curve for a column and explain the salient points on it. 13.10 A column is subject to a uniaxially eccentric load which results in a point (on the interaction diagram) that lies (i) marginally outside (ii) marginally inside the envelo~eof the 'design interadion curve'. Comment on the safety of the column for the two situations. 13.11 Explain the reinforcement arrangement details underlying the design interaction curve given in SP : 16'for the condition "rectangular section with reinforcement dis&buted equally on four sides". 13.12 Briefly explain the difficulties in a rigorous analysis for the design strength components of a given rectangular column section under biaxial loading. - 13.13 Explain the basis for the simplified ,Code procedure for analysing the design strength components of a biaxially loaded column with rectangular cross section. 13.1

13.14 What is the main difference, in t e r n of structural behaviour, between a 'short column' and a 'slender column'? 13.15 Distinguish between 'member stability effect' and 'lateral drift effect' in slender column behaviour. 13.16 In frame analysis, the columns are assumed to be fixed at their bases and the foundations have to be designed to resist the base moments as well as axial loads. In the case of slender columns located at the lowermost storey, is it necessary to include 'additional moments' (due to slenderness effect) while designing the foundations7 (Hint: Does this depend on whether the frame is 'braced' or 'unbraced'?)

.

*

,.

,.

.

.

13.2

13.4

A seven-storeyed building has a floor-to-floor height of 4m and a plan area!$ 18m x 30m with columns spaced at 61n intervals in the two directions. Assume that all columns have a size 400 mm x 400 mm with M 25 concrete, and all primary beams have a size 250 rmn x 600 mm with M 20 concrete.

13.5 13.6

(b) Determine the effecrive lengrhs of a comer column in the second storev~ ~, ~ . With reference to the short column section shown in Fig. 13.35, assuming axial loading conditions, determine the maximum service load that the column can be safely . subiected to: ~ - . (i) according to the LSM provisions of the Code (assuming a load factom of 1.5) ~, .\ (ii) according to the WSM provisions of the Code ~

- .

13.7 13.8

~

13.9

@

Fig.13.37 Problem 13.6

Design the reinforcement in a column of size 400 mm x 600 mm, subject to a factored axial load of 2500 k N The column has an unsupported length of 3.0 m and is braced against sideway in both directions. Use M 20 concrete . and Fe 415 steel. Repeat Problem 13.4, co~~sidering a circular colu~nnof 400 nun diameter. Assume (i) lateral ties (ii) spiral reinforcement. For the column section showu in Fig. 13.37, determine the design strength components correspondi~lgto (i) the condition of 'bnlanccd iailure'; (ii) x , , / D = 0.55; (iii) x , , / D = 1.1. Assume bending wit11 respcct to the major axis. Repeat Problem 13.6, considwing bending with respect to the ninor axis. Generate the design interaction curves for the colu~nm~ section in Fig. 13.37, considering uniaxial eccentricity with respect to (i) the major axis Ci) the ~ninoraxis. [It is convenient to achieve this with the help of a suitable computer program]. Verify with reference to the charts in SP : 16. For the L - shaped section shown in Fig. 13.38, determine the desig~lstrength components coresponding to the neutral axis location shown in the Figure. ~

.

.

Flg. 13.36 Problem 13 3

s

(a) Determine the stability indices of the structure in the transverse and longitudinal directions, considering the second storey. Assume a total distributed load of 50 kNlm2 from all the floors above combined.

.

8 4 TIES 200 CIC

clear cover = 40 mm

PROBLEMS 13.1

4-25 P

6 $spiral G ' , 50 mm clc pitc

~

Clear Cover = 40 rnm

i steel I

8 P TIES @ 200 clc

Flg. 13.35 Problem 13.2 13.3

Repeat Problem 13.2 with reference to the column shown in Fig. 13.36. [Hila: The 5 percent increase in strength is allowed subject to certain conditions. Verify].

Fig. 13.38 Problem 13.9

clear cover

652 REINFORCED CONCRETE DESIGN

13.10 A sl~ortsquare colnnm 300 mm x 300 lmn is rcinlorced with 4 bars of 25 $, placed with a clear cover of 45 mm. Assuming M 25 concrete and Fe 415 steel, determine (i) the maximum eccentricity with which a factored load of 1250 kN can be safely applied: (ii) of , , the maximum factorcd load that cah be applied at an ecccl~tricity 400 mm. 13.1 1 A short circular tied colomn 350 nun diameter is reinforced with 6 bars of 20 $, placed with a clcar cover of 40 mm. It is subject to a factored axial load of 1000 kN, combincd with fnctorcd bending moments of 50 kNm each applied in two perpendicular directions. The concretc is of grade M 25 and the steel of grade Fe 415. Check the safety of the column. If the column is found to be unsafe, suggest suitable modification to the proposed reinforcement. 13.12 Design a short squam column, with effective length 3.0n1, capablc of safely resisting the following factored load effects (under uniaxial eccentricity): ri) P..,, = 1625 k ~M.. . = 75 !dim (ii) P,, =365 kN, M,,= 198 kNm. Assume M 25 concrete and Fc 415 steel. 13.13 Repeat Problem 13.12, considering a suitably proportioned rectangular section. 13.14 Repeat Problem13.12, considering a circular column with spiral reinforcement. 13.15 Design the reinlorcement far a column with I , = I , = 3.5m and size 300 mm X -5flfl - - mm. ~ ~subiect ,- to ~ a lactored ~ ~ axial . load of 1250 kN with biaxial moments of 180 kNm, and LOO kNm with respect to the mnjol. axis and pinor axis respectively (i.e., M,, = 180 kNm, M , ,= 100 kNm). Assunre M 25 concrete and Fe 415 steel. 13.16 Reoeat Problem 13.15, considering P,,= 1500 kN, M,, = 100 M, M , , = 80 kNm. 13.17 Consider a square column, 400 nun x 400 mm, with 4 - 25 $ bars at corners placed with a clear cover of 45 mm, and I,, = I,, = 12D, subject to axial loading conditions. Determine the maximum factored axial load P,, that the column can safely carry considering: \-,

~

DESIGN OF COMPRESSION MEMBERS

Assume the column to be braced, and pinned at both ends in both directions. Assume M 25 concrete and Fe 415 steel, and design by (i) strength reduction coefficient method; (ii) additional monient method. 13.19 Repeat Problem 13.18(ii), considering biaxial moments M,,= M , , = 1001d\lm in addition to P,,= 1100 kN. REFERENCES 13.1 - Commentary on Building Code Requirements for Reinforced Concrete ACI 318-95, American Concrete Institute, Detroit, 1995. 13.2 - Structural Use of Concrere: Part 1: Code of Pmctice for Design and Constructioti, BS 8110 : Part 1 : 1997, British Standards Institution, London, 1997 -... .

13.3

13.4 13.5

13.6 13.7

~

(a) short column behaviour under axial loading, assuming I = 120, (b) short column behaviour under biaxial loading with mininlum

13.8 13.9 13.10 13.11

13.12

eccentricities; (c) , . slender colulmi behaviour (considering 'additional eccentricitics' alonc').

Comment on the results obtained. Assume M 20 concrete and FC 415 stcel. 13.18 Design thc reinforcement in a column of sizc 250 tnm x 400 nun. with an unsopportcd length of 6.0 m, subject to a facto@ axial load of 1100 kN.

' That is, assuming zero primary inomenrs

653

13.13 13.14 13.15

Wood, R.H., EJfrcrive Lengths of Columns in Multi-Storey Buildings, The Structnral Engineer, Vol. 57, Nos 7-9 (3 parts), July, August and Septelnber 1974 -. . Kavanagh, T.C., Effective Length of Framed Columns, Transactions, ASCE, Vol. 127, Part 11, 1962,... oo 81-101. Breen J.E., MacGregor, J.G., and Pfrang, E.O., Deteminntiorz of Effective Length Factom for Slender Concrete Columns, Journal ACI, Vol. 69, No. 11. Nov.1972, pp 669-672. Timoshenko, S.P. and Gere, J.M., Theory of Elastic Srabilify, Second edition, McGraw Hill International edition, 1963. -Explanatory Handbook on Indian Standnr.d Code of P~xcticcfor Plain and Reinforced Concrete (IS 45619781, Special Publication SP:24, Bureau of Indian Standards, New Delhi, 1983. Taranath, B.S., Strrtctural Analysis and Design of Tall Buildings, .McGmwHill International cdition, 1988. Bresler, B. and Gilbert, P.H., The Requiwnents for Rei?rforced Concrete Colusms, Journal ACI, Vol. 58, No. 5, November 1961, pp 555-570. Park, R. and Paulay, T., Reinforced Concrete S w u c r u ~ sJohn , Wiley & Sons, Inc., New York, 1975. Rao, P.S and Menon, D., Ultimate Strerrgth of Tubular Reirlforced Concrete Tower Sections Under Wind Loading, Indian Concrete Journal, Februa-y 1995, pp 117-123. - Design Aids (for Reinforced Concrete) to IS 4 5 6 : 1 9 7 8 , Special Publication SP:16, Bureau of Indian Standards, New Delhi, 1980. Press, W.H., Flannery, B.P., Teukolsky, S.A. and Vetterling, W.T., Nurner.icirl Recipes in C, C:nnbridge University Press, Cambridge, 1988. Bresler, B., Design Criteria for Reinforred Concrete Colu,nws Under Axial Load mzdBiuxia1 Bending, Joumal ACI, Vol. 57, 1960, pp 481490. Gouwens, A.J., Biaxial Bending Simplified, Reinforced Concrete Cola,nrts, ACI Special Publication SP-50, American Concrete Institute. Detroit. 1975. pp 223-261.

DESIGN OFFOOTINGS AND the soil strata on top of which the substructure is to bc iounded. This comes under the specialised domain of geotechnical engineering (soil mcchnnics), and lor important struclures andlor difficult soil conditions, thc type of foundation to be used is based on a soil s t ~ ~ dbyy a geotech~ucalconsultant. In the casc of rctaining walls, the choice of thc type of wall is governed by the height of the cart11 to be mtained and other sitelsoil conditions. It is not the objective of this book to covel' the designs of all thc different typcs of foundations and retaining walls. Nor is it thc objective of the Code on Reinforced Concrete Design (IS 456) to do this. The Code recommendations (CI. 34) are confined to the design of footings that support isolated cohrn~nso r walls and rest directly on soil or on a group of piles [Ref. 14.11. This chapter is, accordingly, confined to the design of thesc simple types of footings' (including combined footings suppo;ting two columns) as well as retaining walls (carrtileeer. and corrnterfort walls). ; Thesc simple types of footings [Fig. 14.11 are the most widcly used types of foundation and are relatively cheap to build. The dcsign o r more coml>lextypes of foundations (continr~ous footings, raft foundations, pile ioundations, wclls and caissons, etc.) is clearly outside the scope of this book, and for this, reference may be made to book? on foundation engineering [Ref. 14.2, !4.3J and related IS Codcs [IS 2911 (Parts I-1111, IS 2950, etc.]. The special codes ~clatetlto the design of simple footings (discussed in this chapter) are IS 1904:1986 [Ref. 14.41 and IS 1080:1980 [Ref. 14.51. Sections 14.2 - 14.6 deal with the types, behaviour and dcsign of footings, while Sections 14.7 - 14.9 deal with the types, behaviour and design of retaining walls.

if & FLAT

RETAINING WALLS 657

STEPPED

SLOPED

(a) isdiated footings

@ central beam (if required)

14.2 TYPES O F FOOTINGS 'Footings' belong to the category of shallow fo~rnr(orio,rs (as opposed to deep four~rlnriorrssuch as pilcs and caissons) and are used whcn soil of sufficient strength is available within a relatively short depth below the ground surface. Shallow foundations co~nprisenot only foorings (which support columnslwalls, and have a limited arealwidth in plan) but also rafts which support multiplc columns on a large plan area). The shallow foundation (footing or raft) has ;I large plan area in comparison with the cross-sectional area of the column(s) it supports because:

(b) combined footings

the loads on the colunu~s(axial thrust, bending momentsr) are resisted by concrete under compression and reinforcing steel under tension andlor compression, whereas these load effects are traiamitted by the footinglraft to a relatively weak supporting soil by bemjng pressures alone; the 'safe bearing capacity' of the soil is very low (100 - 400 kPa) in comparison with tile ycr.,nissible comp~essivcs t r e w s in concretc (5-15 MPa) and stecl (130190 MPa) in a column under servicc loads. -

'The design of pile caps is not included in this chapter. Sheiu forces are also induced in columns, which may result in significant horizontal forces at column bases, under latcral loads. These are resisted by friction between the underside of the footing and the soil below, and also by passive resistance of thc soil adjoining the sides of the footing, and in same cases, by 'keys' cast integrally with the footing.

'

( c ) strap footing

(d) wall footing

Fig. 14.1 Types of footings

14.2.1 Isolated Footings For ordinary structures located on reasonably firm soil, it usually suffices to provide a separate footing for every column. Such a footing is called an isolated footing. It is generally square or rectangular in plan: other shapes are resorted to under special circumstances. The footing basically comprises a thick slab which may be flat (of uniform thickness), stepped or sloped (on the upper surface), as shown in Fig. 14.l(a). The soil bearing pressures from below tend to make the base slab of the footing bend upwards, somewhat into a saucer-like shape (cantilever action), and hence the footing needs to be suitably reinforced by a mesh provided at the bottom of the slab. However, in the exceptional case of very small aud relatively thick footings, the structural action is likely to occur, not by bending of the footing slab, but by a lateral dispersion of the compressive stress at the base of the column; in such a case, it suffices to provide a plain concrete pedestal footing [refer Section 14.4.71, The term 'pedestal' is also used to refer to that portion of a column below ground level where the cross-sectional dimensions are enlarged. The provision of a pedestal is optional, but is often resorted to by design engineers, as it rcsults in reduced development length requirements for the colunm bars, reduced slenderness of the column' (especially when the founding depth is large), increased direct bearing area on thc footing base slab, and reduced shear stresses and design moments. Pedestals are also used to support structural steel columns, the load transfer between the steel column and the concrete pedestal being achieved generally through gussetted steel base plates with 'holding down' bolts. 14.2.2 C o m b i n e d F o o t i n g s In some cases it may be inconvenient to provide separate isolated footings for columns (or walls) on account of inadequate areas available in plan. This may occur when two or more columns (or walls) are located close to each other andlor if they are relatively heavily loaded andlor rest on soil with low safe bearing capacity, resulting in an overlap of areas if isolated footings are attempted. In such cases, it is advantageous to provide a single corrrbined footing [Fig. 14.l(b)l for thc columns. Often, the term 'combined footing' is used when nvo columns are supported by a common footing, the term 'conti~iuousstrip footing' is used if the columns (three o r more in number) are aligned in one ditectio~~ alone, and the term 'raft foundation' ('mat foundation') is used when there is a grid of multiple columns'. The combining of footings contributes to improved integral behaviour of the structure. Fig. 14.l(b) also shows a two-Colunm combined footing, in which there is a 'property line' which restricts the extension of the footing on one side. In this case, the non-availability of space near the exterior column is circumvented by combining

'

Tie beams are also solnetimes provided (for this purpose), interconnecting different calums at the lop ofpedestal level (about 150 nun below ground level). Plinth beams also serve as tie beams. Tie isft foundation consists of a thick slab which may be (i) of uniform thich~ess(flat plate), (ii) with locally thicker panels near coiumi bases (flat slab), or (iii) with stiffening beams interconnecting the columns.

the footing with that oC ao interior column. The width of the footing may be kept unifom or tapered, as shown. The trapezoidal shaped footing (with a larger width near the extel-ior columml) is r6quired when the exterior column is more heavily loaded than the interior column. Another option is a combined fooling which is T-shaped. It is sometimes cconomical to providc a central beam interconnecting the column bascs; this causes the base slab lo be~ld transversely, while the bean] alone bends longitudinally. An alternative to the conventional combined footing is the strop footing, in ivbich the columns are supported esscntiaily on isolated footings, but interconnected witti a beam, as shown in Fig. 14.l(c). 14.2.3 Wall F o o t i n g s Reinforced concrete footings are required to support rcinforccd concrete walls, and are also sometinles employed lo support load-bearing masonry wallst. Wall footings distribute the load from the wall to a wider area, and are continuous thmughout the length of the wall [Fig. 14.1(d)]. The footing slab bends essentially in the direction transverse to the wall (a 'one-way' slab), and hence is reinforced mainly in the transverse direction, with only distribulors in the longitudinal direction. 14.3 SOIL PRESSURES UNDER ISOLATED FOOTINGS 14.3.1 Allowable Soil P r e s s u r e The plan area of a footing base s l a b i s selected so as to limit the maximum soil bearing pressure induced below the footing to within a safe limit. This safe limit to the soil pressure is determined using the principles of soil mechanics [Ref. 14.2, 14.31. The main considerations in detcrnuning the allowable soil pressure, as well as fixing the depth of foundation, are (i) that the soil does not fail under the applied loads, and (ii) that the sctllemen~s,both overall and differential, nre within the limits permissible for the structure. Tile safety factor, used in soil mncchanics, lies in the range 2 - 6, and depends on the type of soil, and related uncertainties and approximnations. ' It should be noted that tilc value of the safe soil b e a ~ b t gcupociry ('allowable soil pres'sure'), q,,, given to the stlllctur~ldesigner by thc geotechllical c o ~ l ~ l l ~ t a is llt~, applicable for suvice loud conditions, as q, includes the factor of safety. Hence, thc calculation for the required area of CI footing must be based on q , and the suvice load effects. The 'partial load facton' to be used for different load combinations (DL, LL. WLIEL) should, therefore, bc thosc applicable for the seruicenbiliry linlit store and rror the 'ultimate limit state' [refer Scction 3.631 when uscd in association with q,,. Arrother point to bc noted is that ihe prescribed allowable soil pressure q, at a given depth is generally h e g'vss pressure, which includcs the pressme clue to the existing overburdell (soil up to the founding depth), and not the ner pressure (in excess of the existing ovcrburdcn prcssore). Hence, the total load to be considered in 'It is inox common to have stepped [masonry (stone or brick) foundation far nmonry walls. The soil bearing capacity, according to soil mechanics thcory, dqleods aa the size of the footing, and this is to be nccounrcd for (approximately) in the reconunend.ltiol1dai made in the Soil Report.

'

DESIGN OF FOOTINGS AND RETAINING WALLS 661

660 REINFORCED CONCRETE DESIGN

calculating the maximum soil pressure q (2 q,) must include the weight of the footing itself and that of the backfill. Often, in preliminary calculations tbese weights we accounted for approximately as 10 - 15 percent of the axial load on the column; however, this assumption should be verified subsequently.

14.3.2 Distribution of Base Pressure The distribution of the soil reaction acting at the base of the rooting depends on the rigidity of the footing as well as the properties of the soil. The distribution of soil pressure is generally non-oniform. However, for convenience, a linear distribution of soil pressure is assumed in normal design practice.

Eccentrically Loaded Footings The load P acting on a footing may act eccentrically with respect to the centroid of the footing base. This eccentricity e may result from one or more of the following the column transmitting a rnomcnt M in addition to the vertical load [Fig. 14,3(a)]; the column carrying a vertical load offset with respect to the centroid of the footing [Fig. 14.3(b)J;

Concentrlcally Loaded Footings Thus, in a symmetrically loaded footing, where the resultant vertical (service) load P + AP (where P is the load from the column and AP the weight of footing plus backfill) passes through the centroid of the footing, the soil pressure is assumed to be uniformly distributed [Fig. 14.21,and its magnitude q is given by

q=- P + A P A wherc A is the base area of the footing.

1' (b)

....................

GROSS SOIL PRESSURE

'__1 8

resultant thrust

Y

It--------L

a.......................

......................

area A = EL

I

I Y

.

Fig. 14.3 Eccentric loading on a footing

,

n a general casc, biaxial eccentricities (i.e., eccentricities of loading with respect

Flg. 14.2 Assumed uniform base pressure distribution under concentric lea Limiting q to thc allowable soil pressure (/n will give [lie nunimum requked fooling:

P+M Arc,#

=z

p

90

elltricities i n loading can be quite significantin footings which support colu~msthat form rtof a lateral load resisting frame. However, as the lateral lands are generally assumed to act

o E s l ~OF~FOOTINGS 662

AND RETAINING WALLS

663

REIN F O R C E D CONCRETE DESIGN

il under compression alOnc), a'!d soil reaction R with the ecce'ltrtc

For the purpose of determining the base pressures under eccentric lo footing is assumed to be rigid and the contact pressure distribution to b magnitude of the pressure distribution is determined from consideratio static equilibrium. Essentially this means that the centre of pressure (th the resultant soil reaction R acts) must be collinear with the resultant lin the eccentrically applied load P + AP,with R = P + dP [Fig. 14.41. For preliminary calculations, AP, the weight of footing plus backfill, ma as 10-15 percent of P. Various possible linear base pressure distlib depicted in Fig. 14.4 for the case of uniaxially eccentric loading on a r footing.

(14.3)

Casel: ( e ( < L / 6 If the resultant loading eccentricity e = MI(P + AP) lies within the "middle the footing (i.e., I e ( 5L/6), it is seen that the entire contact area of the subject to a (nonunifom) pressure which varies linearly front q,,,,, [Fig. 14.4(a)l. These pressures are easily obtained by superposing the separate ef duc tothe dircct load (P + AP) and the bending moment M = (P + dP) e:

. . /c--------L ----+I

(P+AP) (P+AP)e 4.ax.rmn = ---- i' A Z with area A = BL and section modulus Z = ~ ~ ' 1 6wllere , L is the lengtll of tile footi ill the direction of the eccentricity e, and B the width of the footing. Accordiagly, for

1 e 1 5 L/6

(b) e > U6

(14.2

I /

In the limiting case of e = L / 6 . q,,,,= 0 and q,,,, = 2(P + AP)IA, resulting in triangular pressure distribution. The uniform pressure distribution q = ( p + &) [Eq. 14.11 is obtained as special case of Eq. 14.2b, with e = 0. This limiting case of e = L/6, is valid only for uniaxial bending. In caseof

-

$q=(~+~P)/(BL)

/ I

axial bending, the limiting case shall be taken as (c) e=O

e,+L ~ / 6

L' = 3c c=0.5L-e

When the resultant eccentricity e exceeds Ll6, Eq. 14.2 becomes invalicl bccause will.yield a negative valuc for q,,,,, implying a tensile force at the interface. How

One at a lime.

' It1 fact, it can be expected that the soil will tend to separate from the fooling base, tllereby offering no presswe whatsoeve!. in the base regions fatthest removed from q,,,*-.

base pressure distributions under uniaxiallY eccentric loading on rectangular footings (14.4)

Thus, it is seell that the effective length of contact is reduced from to L'= 3c3 q,,,, is increased from ( P + M)IA lo twice the load nd, the soil by tile effective area BL' . In order to limit h:to the ($ divided

+

664 REIN F O R C E D

DESIGN

CO NC RETE DE S IG N

bearing pressure q,,. and also to maxhnise the effcctivc bearing area ratio r / L , it

C a s e 3: Eliminating Eccentrlclty In L o a d i n g Where the magnitude of eccentricity in loading is known with some degree o solution by laterally shifting the footing base, relativc to the column, such that the effective eccentricitv in loadine is reduced considerablv. if not eliminated altogether., . This is not only desirable from the viewpoint of economy but also desirable from the viewpoint of eliminating possible titling of the footing on account of non-uniform: base pressure. The ideal situation of zero effective eccentricity is depicted in Fig. 14.4(c), where it is shown that by suitably offsetting the footing base so that the resultant line of thrust passes through thc ccntroid of thc footing, a uniform pressure distribution is obtainable, with q = (P + AP)lA. Howcvcr, some incrcasc in bearing prcswrc should be considered in practice, to account ibr possible variations in the estimated MIP ratio. Indeed, such a design solution becomes impracticable wllcn the MIP ratio is highly uncertain in magnitude, and especially when the bending moment can be reversible (as under wind loads).

-

OF FOOTINGS AND RETAINING

WALLS 665

against sliding is obtained by friction between the concrete footing base and the soil below, as well as the passive resistance of the soil in contact with the vertical faces of the footing. Improved resistance against sliding can be obtained by providin g a local 'shear key' at the base of the footing, as is sometimes done in foundations for retaining walls. Such a 'shear key' serving as consrruclion joint, may also be provided at the intcrface of the wall/column and the footing, thereby facilitating the transfer of horizontal shear forces (due to lateral loads) at the base of the wnll/colotnn. The restoring moment, counterbalancing the overturning nloment due to lateralleccentric loads is generally derived from the weight of the footiug plus backfill. In some cases, this may call for footings with large base area [refer Fig. 14.4(b)l and large depths of foundation. However, in cases where the overturning moment (not due to wind or earthquake) is not reversible, the problem can be more economically solved by suitably making the column/wall eccentric to the centre of the footing [refer Fig. 14.4(c)l. Another possibility, rela&ely rare in practice, is the case of pullour of a foundation supporting a tension member. Such a situation is encountered, for example, in an overhead tank (or silo) structure (supported on multiple columns), subjected to a very severe lateral wind load. Under minimal gravity load conditions (tank empty), the windward columns are likely to be under tension, with the result that the forces acting on these column foundations will tend to pull out the column-footing from the soil. The counteracting forces, comprising the self weight of the footing and the weight of the overburden, should be sufficiently lwge to prcvent such a 'pullout'. If the tensile forces are excessive, il may be necessary to resort to tension piles for proper anchorage. 14.4 GENERAL DESIGN CONSIDERATIONS AND CODE REQUIREMENTS 4.4.1 Factored Soll P r e s s u r e at Ultimate Limit S t a t e

14.3.3 Instability Problems: Overturning a n d Slldlng When lateral loads act on a structure, adequate stability of the structure as a whole should be ensured at the foundation level - against the possibilities of overturning and sliding. Instability due to overluming may also occur due to eccentric loads, in footings for columns which support cantilevered beams/slabs. The Code (CI. 20) recommends a factor of safety of not less rhan 1.4 against both sliding and overturningt under the most adverse combination of the applied characte~isricloads. In cases where dead loads contribute to improved safety,, i.e., increased frictional resistance against sliding or increased restoring moment against overturning moment, only 0.9 times the characteristic dead load should be considered. It may be noted that problems of overturning and sliding arc relatively rare in reinforced concrete buildings, but are commonly encountered in such structures as retaining walls [refer Section 14.81, chimneys, industrial sheds, etc. The resistant

As mentioned earlier, the area of a footing is fixed on the basis of the allowable bearing pressure q, and the applied loads and moments under service load conditions vith partial load fflcrors applicable for the 'serviceability limit statet'). Oncc the ase area of the footing is determined, the subsequent structural design of the footing done for the factored loads, using the partial load factors applicable for the ltimate limit state'. In order to compute the factored moments, shears, etc., acting at a1 sections of the footing, a fictitious factored soil pressure q,,, corresponding to e factored loads, should be considered. It may further be noted that the soil pressure which induces moments and shears in footing base slab are due to the net pressure q,,,,, i.e.. excluding the pressure nced by the weight AP of the footing and the backfill (assumed to be uniformly distributed). This net pressure is due to the concentrated load on the column (from

~ (Cl. 20.1) permits a reduced n~nirnurnfactor of safety of I Against overturning, t l Code the overturning moment is entirely due to dead loads. However, it is advisable to ap unifonn mini~numfactor safety of 1.4 in all cases of loading.

As mentioned in Section 3.6.3, the partial load factor may be taken as unity in general except for the load combination DL + LL + WUEL, where a partial load factor of 0.8 is applicable for live loads (LL) and for wind loads (WL)learthqu&e loads (EL).

'

666 REINFORCED CONCRETE DESIGN

the superstmcture) and the moments at the base of the colu~lu~ (or pedestal), as shown in Fig. 14.5. Using g,vssp,rssurps instead of nerpressures will result in needlessly'' conservative designs. The 'factored net soil pressure' q,, to be considered in tl~e design of the footing at the limit state is obtainable from the factored loads on the column (P,,, M,,)as shown in Fig. 14.5(b). ,,, 't

P

from the colum~/pedestalto the footing, and in cases where horizontal forces are involved, safety against sliding and ovcrturning. Deflection control is not a consideration in the design of footings which are buried underground (and hence not visible). However, control of crack-width and protection of ~inforcenlentby adequate cover are important serviceability consider8tions, particularly in aggressive environmnents. It is considered sufficient to limit the crackwidth to 0.3 mm in a majotity of footings, and for this the general detailing requircments will serve the purpose of crack-width control [Ref. 14.11.

14.4.3 T h i c k n e s s of Footing B a s e S l a b

P

M

net soil

A

z

pressure

T$++

The thicla~essof a footing basc slab is generally based on considerations of shear and flcxure, which are critical near thc colu~ru~ location. Generally, shear considerations predominate, and the thickness is based on shear criteria. Except in the case of small footings, it is economical tu vary thc thickness from a minimum at the edge to a maximum near the face of the @.Olu~ium, in keeping with the variations in bending moment and shear force. This may be achicved either by sloping the top face of the base slab or by providing a stepped footing. In any case, the Code (CI. 34.1.2) restricts t!le minimum thickness at the edge of the footing to 150 nun for footings in general (and to 300 mm in the case of pile caps). This is done to ensure that the footing has sufficient rigidity to provide the calculated beaing pressures. A 'levelling course' of lean concrete (about 100 nun thick) is usually pmvided below the footing base.

Y

(b)

Fig. 14.5 Net soil pressure causing stresses in a footing

14.4.2 General Deslgn Considerations The major design considerations in the structural design of a footing relate toflexulexure, shear (both one-way and two-way action), bearing and bond (development length). In these aspects, the design procedures are similar to those for beams and two-way slabs supported on columns. Additional considerations involve the transfer of force

14.4.4 Design f o r S h e a r The thickness (depth) of the footing base slab is most often dictated by the need to check, shear stress, and for this reason, the design for shear usually preccdes the design for flexure. Both one-way shear and two-way shear ('punching shear') need to be considered in general [refer CI. 34.2.4.1 of the Code]. However, in wall footings [Fig. 14.l(d)l and combined footings provided with a central beam [Fig. 14,l(b)], the base slab is subjected to one-way bending, and for this reason, need to be designed for one-way shear alone. The critical section for one-way shear is taken, as for beams, at a distance d (effective depth) from the face of the coludpedestal [Fig. 14.6(a)l or wallheam [Fig. 14.6(d)]. The effective area resisting one-way shear [Fig. 14.6(a), (d)] may be rectangular or polygonal, depending on whether the footing is flat [Fig. 14.6(a)] or sloped [Fig. 14.6(c)].

DESIGN OF FOOTINGS AND RETAINING

668 REINFORCED CONCRETE DESIGN

Critical seclions for moment

'

h i t i c a l secllon for

'

around) for two-

way shear Vu2

,,crilical seclians for moment

maso

wall

(c)

Fig. 14.6 Critical sections for shear and moment

The behaviour of footings in two-way (punching) shear is identical to that of a two-way flat slab supported on columns, discussed in Chapter 11. The critical section for two-way shear is taken at a distance dl2 from the periphery of the column, as shown in Fig. 14.6(b), (c). The design procedures for one-way and two-way shear are identical to those discussed in Chapters 6 and 11 respectively. However, shear reinforcerncnt is generally avoided it1 footing slabs, and the factored shear force V,, is kept below the factored shear resistance of the concrete v,,' by providing the necessary depth. Where. for some reason. there is a restl.iction on t l ~ edeoth of the footing base slab on account of which V,. > V,,c,appropriate shear reinforcement should be designed and provided, to resist the excess shear V,,- V,,,. Finally, it may be noted that in the case of a column/pedestal with a circular or -octagonal cross-section, the Code (CI. 34.2.2) recommends that an equivalent square section should be considered, for the purpose of locating the critical sections for shear (and moment). The equivalent squares should be inscribed within the perimeter of the round or octagonal column or pedestal.

As mentioned carlier, the footing base slab bends upward into a saucer-like shape on account of the net soil pressure q,, from below [Fig. 14.6(a)l. Based on extensive tests, it has been determined that the footing base slab may be designed against flexure by considering thc bending moment at a critical section defined as a straight section passing through

(a)

,

t

WALLS 669

the face of a column, pedestal or wall for a footing supporting a concrete c o l u m ~ , pedestal or wall [Fig. 14.6(a)l: halfway betweenthe face and centreline of the wall for a footing supporting masonry wall [Fig. 14.6(d)]. In one-way reinforced footings (such as wall footings), the flexural reidorcement (calculated for the moment at the critical section) is placed perpendicular to the wall at a uniform spacing. In the perpendicular direction (along the length of the wall), nominal distributor reinforcement should be provided - mainly to account for secondary moments due to Poisson effect and possible differential settlement, and also to take care of shrinkage and temoerature effects. In two-way reinforced square footings also, flexural reinforcement may be placed at a unifo~mspacing in both directions. I n two-way reinforced rectangular footings, the reinforcement in the long direction is uniformly spaced across the full width of the footing, but in the short direction, the Code (Cl. 34.3.1~) requires a larger concentration of reinforcement to be provided within a central band width, equal to the width B of the footing:

-

' POI the purpose of calculating the design shear strength

z, of concrete, a nominal percentage of flexural tensile reinforcement (p, = 0.25) may be assumed (in preliminary alculations).

670 REINFORCED

CONCRETE

DESIGN DESIGN OF FOOTINGS AND RETA!NING

2

Reinforcement in central band width r A,,,,,,,, xP+l where A,,,,,,, -total flexural reinforcement required in the short direction

(14.5)

I

and fi = ratio of the longside (L) to thephort side (U) of the footing. This reinforcement is to be uniformly distributed within the central band width (equal to width U), and the remainder of the reinforcement distributed uniformly in the outer portions of the footing, as shown in Fig.-14.7. This is done to account (approximately) for the observed variation of the transverse bending tnonlent along the length of the footing.

uniform

!

spacing laterally

; -

central band widlh 8

:-

WALLS 671

These special detailing requirements are strictly intended for footings with oniforln slab thicla~ess. In the case of sloped footings, it usually suffices to providc ulliforldy distributed reinforcement in the s l ~ o direction ~t also, as the reduccd bending momcnt in the outer portions is coupled with reduced cffective dcpths in these i-egions. In the long direction also, the common practice is to provide uniforlnly spaced reinforcement througl~outthe width or the footing, despite the variations in depth. In general, the percentage flexural reinforccmnent requirement in footing base slabs is low, owing to the relatively large thickness provided on shear considerations. At any rate, the reinforcement should not be less than the minilnum prescribed for slabs [refer Chapter 51, unless the footing is designed as a plain concrete (pedestal) footing. Furthermore, the percentage reinforcement provided should be adequate to mobilise the required one-way shear strength it1 concxete. It is advisable to select stnall bar diameters with small spacings, in order to reduce crackwidths and development length requiremnents. Devcloptnent length requirements for flexural reinforcement in a footing should bc satisfied at the sections of maxinrum moment, and also at othcr sections where the depth is altered. Shortfall in requircd development length can be made up by bending up the bars near the edges 01the footings. This may be required in footings with small plan din~ensions. Furthermore, the longitudinal reinforcement in the column/pedestal tnust also have the required development length, nreasured From the interface between the colu~dpedestaland the footing. When the column is subjected to cotnpression alone (without the bars being subjcct to lension), it is possible to achievea full transfer 01 forces from the colunuVpcdesta1 to the footing by bearing, as dis,cussed ill the next section (Section 14.4.7). Whcre this is not possible, and the transfer of force is accotnplislted by ~einrorcement, such reinforcen~ent must also have adequate developrnetlt length on cach side.

14.4.6 Transfer of F o r c e s at Column B a s e All forces (axial force, monlent) acting at the base of the colutnn' (or pedestal) must be transferred to the footing either by compression in concrcte or by tension/compression in reinforcing steel. The force transfer achieved through conlpression in concrete a1 thc interface is linlited by the bearing resistance of conc~.etefor either surface (is., supported surface or supporting surrace). Under factored loads, the maximum bearing stress& ,, is limited by the Code (CI. 34.4) to .fhr.,nax

S,

SECTION

'XX'

Fig. 14.7 Detailing of flexural reinforcement In a rectangular footing with uniform thickness

= 0.45fck

(14.6)

where A2 is the loaded area at thc colutnn base, and A , lhc maxilnum area of the portion of the supportbtg surlace that is geon~etricallysimilar to and concentric wit11 the loaded area. In thc case 01stcpped or sloping footings, the area A , is to be takcn as that of thc lower base of ihc largest frustrum of a pyramid (or cone) contained wholly within the footing (wit11area Az on top) wit11 a side slope of 1 in 2, as shown in 'This is equally applicable in the case of force transfer from the column base to the pedestal (if

provided) and frotn the pedestal bose to the footing.

672 REINFORCED CONCRETE DESIGN

DESIGN OF FOOTINGS AND RETAINING WALLS

Fig. 14.8(a). Thc factor m i n Eq. 14.6 allows for the increase in concrete strength in the bearing area in the footing due to confii~ementoffered by the surrounding concrete. This factor is limitcd to 2.0. A limitation on the bearing stress is imposed because very high axial compressive stresses give rise to transverse tensile strains which may lead to spalling, laeral splitting or bursting of concrete. This possibility, howcver, can be countered by providing suitable transverse and confincment reinforcement.

Fig. 14.8(b). The diameter of dowels should not exceed the diameter of the colunm bars by 3 mm.Furthermore, the reinforcenlent provided across the interface rllust comprise at least four bars, with a total area not less than 0.5 percent of the crosssectional area of the supported column or pedestal [refer C1. 34.4.3 of the Code]. Finally, it should be ensured that all reinforcement provided across the interface (whethcr by extension of column bars or dowels) must have the necessary development length in compression or tension, (as applicable) on both sides of the interface. Whcre pedestals are provided, and full force transfer is possible at thc interface of column and pedestal, no reinforcement is theoretically required in the pedestal. However, the Code (CI. 26.5.3.1h) specifies that nominal longitudinal reinforcement (i.e., in a direction parallel to the column load) of not less than 0.15 percent of the cross-sectional area should be provided, for reasons similar to those pertaining to minimum reinforcement in columns [refer Section 13.3.31.

14.4.7 Plain Concrete Footings When the column is relatively lightly loaded (without any bars in tension) and the base area requirement of a footing is relativcly low, it may be economical to provide a simple plain concretc block as a footing. Such a footing is sometimes called a pedestnl footing. If thc bearing stress at the column base under ultimate loads is less than f,,,,>j,, (given by Eq. 14.6), the force transfer from the colimnm base to the footing (pedestal) is achievable without the need for any reinforcement at the interface. Further, if the base area of the footing falls within a certain zone of dispersion of internal pressure in die footing, the entire force is transmitted to the footing base by compression' (sfr-trr action, as shown in Fig. 14.9~).and the soil pressure docs not inducc any bending in the footing. The (inlagindry) struts are inclined to the vertical, and the horizontal compo~lentof the strut forces will necessarily call for some tie action ('strul and tic' concept - see Section 17.2), as shown in Fig. 14.9(b). To carry the tie forces and to avoid possible cracking of concrete duc to the resulting tensilc forces, it is necessary to provide some minimum reinforcement to serve as effective ties [Fig. 14.9(b)].

(a) definition of area A, (GI. 34.4 of Code)

(b) reinforcement across column-footing interface

673

dowd bar

Fig. 14.8 Transfer of forces at column base

It should be noted that h,: , may be governed by tlic bearing resistance of the concrete in the colonm at the interface (for which is obviously unity), rather than that of the concrete in lhe footing (for which 1 i < 2). If the actual then thc excess force is transferred by compressive stress excceds f,,,,,,,, rcinfo~cemcn;,dowels or mechanical connectors. For transferring nmoment at the coluinn base (involving re~lsionin thc reinforcement), it niny be nccessary to provide thc same aniouot of reinforcement in the footing as in thc column, although some relief in ihc compression rei!iforcement is obtainable on account of transfer through baring. This may be achieved by either continuing the column/pcdestnl bars into the footing or by providing scpnrnm rlo\vel bars across the interface as depicted in

For the purpose of defining this zone of dispersion of internal pressure in the footing, thereby enabling the determination of the required thickness of the footing block, the Code (Cl. 34.1.3) defines an angle a between the plane through the bottom edge of tlic footing and thc co~respondingedge of the colunm at the interface [Fig. 14.91, such that t a n u 2 0.9,/100~,,,,

/f,, + 1

(14.7)

where q,,,, is the rnaximutn soil pressure under service loads, as defined earlier [Eq. 14 Za, Fig. 14.4aI.

-

'This is only a convenient idealisation; the actual slate of stress is difficult 10 assess.

DESIGN OF FOOTINGS A N D RE T A I N I N G WALLS

.

- . -. -

-. -..

675

nsfer of axial force a t hase of colmnn lain concrete block footing, full force transfer must be possible at the column base, without the need for reinforcement at the interface. That is, the factored axial load P,,must be less than the limiting bearing resistance Assuming a load factor of 1.5, P,, = 330 x 1.5 = 495 kN Limiting bearing stress&,.,,, = 0.45Lx At the column-footing interface, fb, ,,,,, will be governed by the column face in this case (and not the footing face), with Al = A2 = (300 x 300) m1n2 ~Fb,=0.45x20x3002=810x103~ > P,,= 495 kN

ithout the need for reinforcement. footing + backfill to comprise 10 percent of the axial 330 x 1.1 - 1.01 m2 load, base area required = -----360 Provide I m x l m footing, as shown in Fig. 4.10. (8) Fig. 14.9 Plain concrete (pedestal) footing where t a n a 2 0.9J100q ,,,

/fck

+1

An expression for thc thickless D of the footing block is obtainable [pig. 14.91 as:

D = ( L - b)(tan a)/2 =1

D 2 0.9~100q,,,,

If,

+ I (L- b ) 1 2

where 6 is the width of the column and the expression for talla in E ~ 14.7 . governs the lninimum thickness of the footing. The design of a plain concrete footing is demonstrated in Exampie 14.1. 14.5 DESIGN EXAMPLES O F ISOLATED AND WALL FOOTINGS EXAMPLE 14.1: Design of a Plain Concrete Footing Design a plain concrete footing for a column, 300 mm x 300 llun, an axial load of 330 kN (under service loads, due to dead and live loads). A~~~~~~an allowable soil bearing pressure of 360 liN11n' at a depth of 1.0 En below ground, Assume M 20 concrete and Fe 415 steel.

=, D 2 350 x 0.9 d l 0 0 x 0.36120 + 1 = 527 mm Provide 530 mm. Hence. provide a concrete block 1000 x 1000 x 530 mm. Further, it is necessary to provide minimum reinforcement to provide for 'tie action', and to account for temperature and shrinkage effects: A,,,,,,,, = 0.0012BD = 0.0012 x 1000 x 530 = 636 mm2 Provide 6 - 12 mm bars (A,, = 678 mm2) both ways with a clear cover of 75 mm, as shown in Fig. 4.10. The spacing is within limits (< 5d or 450 mm).

+

Check gross base pressure Assuming unit weight of concrete and soil as 24 kN/m3 and 18 kN/m3 respectively, actual gross soil pressure q , , =~ --~ 330 + (24 x 0.53) + (18 x 0.47) 1.0 x 1.0 = 330.0 + 12.7 + 8.5 = 351.2 kN/m2 - Hence, safe. < q, = 360 kN/ml

.,.

.

. ,.~ , ~ .

,,,

,;::I) t

676

DESIGN OFFOOTINGS AND RETAINING WALLS

REINFORCED CONCRETE DESIGN

.I .J / " . ,,, ,.,

677

Thickness of'footing slab based on shear Net soil pressure at ultimate loads (assuminr a load factor of 1.5)

= 0.383 N/mm2

(a) One-way shear The critical section is at a distanced from the column face [refer Fig. 14.1 11. aFactored shear force V,,,= 0.383 x 3000 x (1275 - 4 = (1464,975 - 11494 N. Assuming z, = 0.36 MPa (for M 20 concrete with, say, p, = 0.25) [reier Table 6.1 or Table 13 of the Code], One-way shear resistance Vcl = 0.36 X 3000 x d = (10804 N V,,,< VCI=, 1464975 - 1149d < 1080d qd2658mm

(b) Two-way shear The critical section is at dl2 from rhe periphery of the column [refer Fig. 14.1 11 JFactored shear force V,,Z= 0.383 x [30002- (450 + 4'1 Assuming d = 658 mm (obtained earlier r ) = 2976.8 x I O ~ N Two-way shear resistance Vr2= k s ~x[4 C x (450 + 4 dl where k, = 1.0 for a square column, and z, = 0 . 2 5 m = 1.118 MPa (refer CI. 31.6.3.1 of the Code) Fig. 14.10 Example 14.1

EXAMPLE 14.2: Square Isolated Footing, Concentrically Loaded

Design an isolated footing for a square column, 450 mm x 450 mm, reinforced with 8-25 g bars. and carrying n service load of 2300 kN. Assmue soil with a safe bearing capacity of 300 kNlm2 at a depth of 1.5 m below groond. Assume M 20 grade concrete and Fe 415 grade steel for the footing, and M 25 concrete and Fe 415 steel for the colunul. SOLUTION

.

Size of footing

= 2300,&I q, = 300 kN/m2 at h = 1.5 m ~ the weight s of uthe footing ~ + backlill ~ to ~ be 10 %' Of the load 2300 = 8.43 p = 2300 kN, base area ~equired= 300

~ l v m P:

~

Millimum size of square footing = Assume a 3 m x 3 m footing base -

'This assumption is verified subsequently.

a 2.904 =

V,, 5 V,, =, 2976.8 x 10) < 2012.4d + 4.472d2 Solving, d 2 621 mm Evidently, in this problem, one-way shear governs the thickness. Assuming a clear cover of 75 mm and 16 41 bars in both directions, with an average d = 658 nun, thichess D 2 658 + 75 + 16 = 749 mm Provide D = 750 mm. The effective depths in the two directions will differ by one bar diameter, which is not significant in relatively deep square footings. For the purpose of flexural reinforcement calculations, an average value of d may be assumed: =,d=750-75-16=659mm Assuming unit weights of concrete and soil as 24 kN/m3 and 18 ~ m respectively, actual gross pressure at footing basc (under senice loads)

'Actual effctive depth provided will not be less than this value: hence, the use of this value in this context can only be on a slightly conseitative side; such an assumption simplifies calculations.

'

678 REINFORCED CONCRETE DESIGN

However, this reinforcement is lcss than assumed for one-way shear design' ( T =~0.36 MPa). for whichp, .i0.25 (for M 20 concrete) a A ,, = 0.25 x 3000 x 6591100 = 4943 tlm? Using 16 mm 4 bars, number of baas required = 49431201 = 25 [corresponding spacing s = (3000 - (75 x 2) - 16)1(25 -1) = 118 mm - is acceptable,] Provide 25 nos 16$ bars both ways as shown in Fig. 14.1 1 Required development length L,, = 0(0'87fy) [refer CI. 26.2.1 of Code] 47, For M 20 concrete and Fe 415 steel, Ld = $ (0.87 x 415)/(4 x 1.2 X 1.6) = 47.0 $ For 16 $ bars in footing, LA~ 4 7 . x0 16 = 752 mm Length available = 1275 - 75 = I200 nun > 752 mm -Hence, OK.

,,,,*,

.. t-l---3000pX

SECTION 'XX'

,, ,

.. ,

section far one-way shear

Transfer of force a t colunm base Factored conlpressivc force at column base: P,, = 2300 X 1.5 = 3450 kN Limiting bearing strcss at column-footing interface, fa, , , = 0.45f,= (i) for column face,f,k = 25 MPa, A, = Az = 4502 mm2 = 0.45 x 25 x 1 = 11.25 MPa (ii) for footing face,f,k = 20 MPa, Al = 3000' mm2, A2 = 450' mm2 a m = 30001450 = 6.67, limited to 2.0

a& ,,..

,

,,

af,,,,,, = 0 . 4 5 x 2 0 x 2 = 18.0 MPa Evidently, the column face governs, and f,,,,>, = 11.25 MPa Limiting bearing resistancc F,, = 11.25 x 450' = 2278.1 x 10'N < P, = 3450 kN *Excess force (to be transferred by reinforcement): AP,,= 3450 - 2278 = 1172 kN This may he transferred by rcinforccment, dowels or mechanical conncctors. In thiscase, it is convenient to extend thc column bars into the footing, as shown in Fig. 14.11. Required developtnent length of the 8-25 $bars provided in thc column, assuming a stress level equal to (0.87fJ x (AP,,IP,,),and M 20 concrete with Fe 415 steel (in compression) 0 (0.87 x 415) For fully stressed bars in compression (M 20, Fe 41.5): Ld = 4(1.2 x 1 . 61.25) ~ = 37.6 4

*

PLAN

Fig. 14.11 Example 14.2

. Design of flexural reinforcement Factored moment at colunu~.face(in either direction): M,,= 0.383 X 3000 X 127S2/2= 933.9 x lo6Nmm + R E - M,, - 9 3 3 . 9 ~ 1 0-~0.717 MPa Bd2 - 3 0 0 0 x 6 5 9 ~-

a

'

Ed =LAXAP,,IP,,

Unless thc fooling dinlcnsions are rwised (to msult i n less shear slrcss). !he reinforcement requirement here will be governed by shear strength requirements, and not tlcxurvl strength requiroeents. If the resultiug I), is excessive, it ,nay be Inore economical to revise the footing dimensions, providing larger plan area and less depth of footing. In a practical design, this should be investigated.

DESIGN OFFOOTINGS AND

680 REINFORCED CONCRETE DESIGN

= 37.6 x 25 x 117213450 = 319 nun. Available vertical embedment length in footing (d = 659 mln) > 319 mm. The bars arc bent (with 90' standard bend) into thc footing, and may rest directly on the top of the reinforcement layer in the footing, as shown in Fig. 14.1 1.

.

Alternntive Design Providing a unifofm thickness of 750 mm for the footing slab is rather uneconomical, as such a high thickness is required esscntially near the face of the column (due to shcar considerations); the effectivc rlcpth requirement falls off with increasing distance from the cricical section for one-way shear; theoretically, only a minimum thickness (150 nun, specified by the Code) need be provided at the edge of the footing. However, the slope provided at the top of the footing should preferably not exceed about 1 in 1.5 (i.c., 1 vertical : 1.5 horizontal), as a stecper slope will require the use of additional formwork on top (to prevent the concrete from sliding down). 660

.I#

450j:

W

4

30 nos 16 $both ways7

.

RETAINING WALLS

681

As the thickness of the footing near the coln~llnbase is governed by shear (one. way shear, in this example) and the effective area available at the critical section is a truncated rectangle, the effective depth required is slightly larger than that for a flat footing. Assuming a thickness D = 750 mm up to a distance of 660 mm (> rl = 659 m) from the periphery of the column; and providing a slope of 1 in 1.5 ovcr the remaining distance of 1275 - 660= 615 mm on all four sides [Fig. 14.121, the edge thickness is obtained as 750 - 61511.5 = 340 mm 3 V,,, = 0.383 X (1275 - 659) X 3000 =707784 N a z,, = 7077841(3000 x 659- 616 X 410)' = 0.410 MPa Providingp, = 0.35 a r , = 0.413 MPa (for M 20 concrete) [refer Table 6.11 z, > z,-Hence, OK. a (A&, = (0.351100) x (3000 x 659 - 616 x 410) = 6036 mmz a No. of 16 $bars required = 60361201 = 30 (as shown in Fig. 14.12). Othcr alternative designs are possible. These include (i) pmviding a proper sloped footing with a thickness varying linearly from a minimum at the edge to a maximunx' at the face of the column, and (ii) providing a stepped footing. In the latter case, the section at the step location becomes a critical scction at which oneway shear, flcxural reinforcement and developrncnt length requirements need to be verified.

EXAMPLE 14.3: Rectangular Isolated Footing, Concentrically Loaded Redesign the footing for the column in Example 14.2, including a spatial restriction of 2.5 m on one of the plan dimensions of the footing. SOLUTION Size of footing As in Example 14.2, requi~edbase area = 8.43 mZ Width B = 2.5 m, a lcngth L = 8.4312.5= 3.37 nl a Provide a rectangular footing 3.4 m x 2.5 In. a Net factored soil pressure = 2300 X lSl(3.4 x 2.5) = 406 k ~ l m ' = 0.406 ~ l m m ' Tlliclu~essrequired for shear An exact solution for the required depth for she= (onc-way shear as well as twoway shear) may be obtained using the conditions V,,, < VCIand < Vc2,as done in Examplc 14.2. In this example, a trial-and-error procedure is used. Assuming an overall depth (thickness) of footing D = 850 mm, with clear cover of 75 mni and 20 nun $ bars in the long direction (placed at bottom) and 16 m m $ bars in the short direction,

PLAN

Fig. 14.12 Example 14.2 -Alternative

''The x e a resisting th 940 mm - OK.

.

(b) s h o ~rlir.ection t (section YY in Fig. 14.13)

I

T

critical section for two-wav shear

critical section lor one-way shear

'X PLAN

Fig. 14.13 Example 14.3

. . . .

M,,,= 0.406 x 3400 x 1025~12= 725.1

X

10' Nmm

his is less than the nlini~nunlreinforcement required for slabs: (A,,),,,,, = o.oo12 b~ = 0.0012 x 3400 x 850 = 3468 mZ Using 16 $ bars, number required = 34681201 = 18 A,, to be provided within a central band width B = 2500 Im is:

--

= 2890 rmn2 -3468x P+l (3512.5 + 1) Using 16 $ bars, number rcqoircd = 28901201 = 15 Provide 15 nos 16 $ bars a1 oniTonn spacing within thc central band of width 2.5 m, and 2 nos 16 $bars each i n the twd outer seglnenls; making a total of 19 bars, as shown in Fig. 14.13. The spacings are within litnits (3d: 300 mm). Required development Icngth = 47.0 X 16 = 752 lllln bevelopnm~tlengtll available = 1025 - 75 = 950 n m ~> 752 111111 - OK.

3468 X

684

REINFORCED CONCRETE DESIGN

DESIGN OF FOOTINGS AND RETAINING WALLS

Transfer of force at column base The calcolations arc identical to those given in Example 14.2 (except that for the footing face, -= 25001450 = 5.56, limited lo 2.0). The excess force of 1171.9 1iN may be transferred across the column-Iooting interface by sinlply extending the column bars, as in the previous Example, and as indicated in Fig. 14.13. Alternative: As in the previous Example, a slopcd footing may be designed; this is Likely to be more economical than a flat footing.

685

7 -10 $distributors

EXAMPLE 14.4: Masonry Wall Footing Design a reinforced concrete footing for a 230 nnn thick masonry wall which supports a load (inclusive of self-weight) of 200 kN/munder service loads. Assume a safe soil bearing capacity of 150 kNlmZat a depth of I m below ground. Assume M 20 grade concrete and Fe 415 grade stcel. SOLUTION Size of footing Given: P = 200 kNlm, q, = 150 kNlmZat a depth of I m. Assuming thc wcight oC the footing + bacldill lo constitute 10 perccnt of the applied load P,and considering a 1 rn length of iooting along the wall, 200 x 1.1 required width of footing = -= 1.47 m. I -SO Provide 1.5 m wide footing.

230 thick masonry wall

1000 wide design strip

PLAN

Fig. 14.14 Example 14.4

~

Thickness of footing based on shear considerations Factureti nct soil prcssurc (assuming a load factor of 1.5) is: 200 x 1.5 y,, = ------ = 200 k ~ l r n '= 0.200 ~ l m m ' 1.5 x 1.0 Thc critical section for (one-way) shear is located at a dislance d away from the face of the wall = V,,= 0.200 x 1000 [(I500 - 2 3 0 ) 1 2 4 = (127000 - 2 0 0 4 N e Assuming nominal flexural reinforcement 0), = 0.25), C' , = 0.36 MPa f01 20 concrete, the shear resistance of concrete is: V,,, = 0.36 x 1000 X d = (360d) N. V,, 5 V,,, =, 127000 - 20Od 5 360d =.d 2 227 1m11 Assuming a clear cover of 75 nvn and 16 @bars, tlliclu~essD 2 227 + 75 + 1612 = 3 10 mm Provide D = 310 n l n ~upto a distance of 250 mm from the face of the wall. At the edge of the footing, a minimurn thickness of 150 mm may bc provided, and tlic thickness h e a d y tapered upto 310 mm, as shown in Fig. 14.14.

.

Design of flexural reinforcenlent The critical section for maximum moment is located halfway between the centrcline and edge of the wall, i.e., at a distance 150012 - 23014 = 692.5 mm from the edge of the footing [refer Fig. 14.41. Considering a I m)ong footing strip with V,,, =852.0!& Hence, one-way shear governs the thickness. As a square footing is provided and the one-way shear requirement is equally applicable in both directions, the d calculated may be taken as an average depth: (d, + dJ2. Assuming 75 mm clear covcr and 12 4 bars, D2357+75+ l2=444mm Provide D = 450 m m and consider the average effective depth, d = 450 - 75 - 12 = 363 mm while designing for flexure. , ,$,

Design of flexural reinforcement Maximum cantilever projection = 845 mm (from face of column) M, = 0.263 x 1950 x 845212 = 183.1 x lo6 W m

.

p, = 0.25 has been assumed for one-way shear strength Accordingly, A,, = 0.25 x 1950 x 3631100 = 1770 mm2 Number of 12 $ bars required = 177011 13 = 16 [corresponding spacing = (1950 - 75 x 2 - 12)/15 = 119 mm -OK]. Provide 16 nos 12 $bars in both directions. m - available [refer Development length required = 47.0 $ = 47.0 x 12 = 564 m Fig. 14,161.

Transfer of forces at c o l ~ ~ nbase m This is as explained in Example 14.5, with the diCferemnce that some of the bars are always under compression, requiring reduced development length. However, the bars in tension need an additional extension of 50 m i beyond the bend point, on account of the reduced footing thickness of 450 nnn (as against 500 mm in Example 14.5). The total extension of 641 + 50 = 741 = 750 tmn requires reorienting the bars diagonally in plan for this length to be available.

14.6 DESIGN O F COMBINED FOOTINGS 14.6.1 General As mentioned in Section 14.2.2, a footing supporting more than a singlc column or wall is called a combined footing, and when many columns (more than two) are iuvolved, terms such as continuous strip footing (if columns are aligned in one direction only) and rafr foundation or mar foundation are used. Multiple column foundations become necessary in soils having very low bearing capacities. However, even in soils havine moderate or hirh 'safe bearina-capacity' . . for the use of individual footings, combined footings become necessary sometinles -as when:

-

WALLS

693

columns are so closely spaced that isolated footings cannot be conveniently provided, as the estimated base areas tend to overlap: an exterior column located along the periphery of the building is so close to the property line that an isolated footing cannot be symmetrically placed without extending beyond the property line. 14.6.2 Distribution o f Soil P r e s s u r e As mentioned earlier (in Section 14.3.2), the prediction of the exact distribution of base pressure under a footing is difficult, as it depends on the rigidity of the footing as well as the properties of the soil. If this is difficult for an isolated footing, indeed, it is more so for a combined footing. For a very rigid footing supported on an elastic soil base, a straight line pressure distribution is appropriate. Such an assumption is found to lead to satisfactory designs in the case of relatively rigid footings. However, for relatively flexible footings, such an assumption is not realistic; the problem is rather complex and involves consideration of soil-structure interaction.

of soil pressure is assumed. 14.6.3 Geometry of Two-Column Combined F o o t i n g s Examples of two-column combined footings are shown in Fig, 14.17. The geomtry of the footing base should preferably he so selected as to ensure that the centroid of the footing area coincides with the resultant of the column loads (including consideration of moments if any, at the column bases). This will rcsult in a uniform distribution of soil pressure, which is desirable in order to avoid possible tilting of the footing (as nicntioned earlier in Section 14.3.2). The footing may be rectangular or trapezoidal in shape [Fig. 14.171, depending on tlie relative magnitudes of loads on the two columns which the footing supports. When the exterior column (which has the space limitation for an independent footing) carries the lighter load ( ? > s / 2 ) , a rectangular footing [Fig. 14.17(b)J or a trapezoidal footing (with a reduced width under the exterior colunmi) as shown in Fig. 14.17(c) may be provided. On the other hand, when the exterior column c a ~ ~ i e s the heavier load [ Z < s / Z i n Fig. 14.17(e)], the wider end of the trapezoidal footing should be located under the exterior column. 14.6.4 Design Considerations in Two- Columns F o o t i n g s

Fixing Plan D i m e n s i o n s As discussed earlier with reference to Fig. 14.17, the plan dimensions of the twocolumn combined footing may be selected to satisfy the following two requirements

D E S I G ~OF FOOTINGS A N D RETAINING

694 REINFORCED CONCRETE DESIGN

WALLS 695

d to be ulliformly distributed' [refer Fig. 14.181.

? alp

shear force

Fig. 14.17 Geometry of two-column combined footings 1.

Base area of footing A =Total (service) loadtIq..

2. The line of action of the resultant of the column loads must pass tlunu centroid of the footing. In the case of a rectangular footing [Fig. 14.17(b)], the second requirement r

Load Transfer Mechanism As in the casc of isolated footings, the factorcd net soil pressure q,, is computed as the resultant lactored load divided by thc base area provided, and the pressure m a ~ b e -

' Including the weight of the footing plus backfill.

laads P,: P2 is subject to uncertainty. Or when these In cases where the of co~urnn by mo,,,ents which may be reversible, the line of action of the loads are of the footing, and the Pressure d'stribution the load will not always match be nonunifom, However,it is col,servativeto assum uniform distnbutlon with lnaximumq,,. t

696 REINFORCED CONCRETE DESIGN

The base slab of the combined footing is subject to two-way bending, and one-way as well as two-way shear (as in the case of isolated footing). In general, the width of the footing (B) is much less than the length (L), with the result that the flexural hehaviour is predominantly one-way (i.e., in the longitudinal direction), and the twoway action (i.e., including transverse bending) is limited to the neighbourhood of the column locations. For the puipose of struct~lraldesign, a simplified (and usually conservative) load transfer mechanism may be assumed - as shown in Fig. 14.18. In this idealised model, thc footing is treated as a uniformly loaded wide longitudinal beam (width 8 , length L and factored load q,,B per unit length), supported on two column strips, which in turn act as transverse beams cantilevered from the columns. The width of each column strip may be taken approximately as the width of the column (a)plus 0.75d on either side of the column [Fig. 14.18(b)J. The thickness of the footing is generally governed by shear considerations, as in isolated foolings. The critical sections for one-way shear are at a distanced from the column face [Fig. 14.18(c)J, and at dl2 from each column periphery for two-way shcar. The distribution of longitudinal shear forces and bending moments may be easily determined from statics, treating the footing slab as being simply supportedT on the two column strips, with overhangs (if any) beyond each column strip, as shown in Fig. 14.18(c), (d). The flexural reinforcement in the longitudinal direction is designed for the 'positive' moment at the face of.the col~lmnand the niaximurn 'ncgative' moment between the columns; the reinforcement is placed at the bottom in the case of the former, and at top in the case of latter, as depictcd in Fig. 14.18(a),(d). The flexural reinforcement in the transverse direction (in the column strip) is designed for the 'positive' moment at the section in line with the face of the columml, considering the column strip as a beam with uniformly distributed factored loads (whose total magnitude is equal to the factored load on the column). This reinforcement is providcd at the bottom, and located in a layer above the longitudinal reinforcement [refer Fig. 14.18(a)l. Nominal transvcrse reinforcement may be provided elsewhere (i.e., other than the column strips), to tic with the longitudinal reinforcement (wherever providcd); these nominal bars, however, are not indicated in Fig. 14.18(a). Dcvclopment length requirements should be satisfied by thc flcxural reinforcement provided. The column strip (transverse beam) should also be checked for one-way shear at a distance, equal to the effective depth of the transverse reinforcement, from the face of the columnlpedestal. The design of a two-column rectangular footing is illusuwed in Example 14.7. Beam-Slab Combined Footings

I i the casc of relatively large footings, providing a uniform large thickness for the entire footing results in a somewhat expensive footing. In such a case, it may be more 'As the design section for both shear and momenf are outside the colomn section, it suffices to assume the supports to be concentrated at the column centrelines; the corresponding shear force and bending moment distributions m shown by dashed lines in Pig. 14.18(c),(d).

DESIGN OF FOOTINGS AND RETAINING

WALLS

697

economical to design a beam-slab footing, in which the footing consists of a base slab stiffened by means of a central longitudinal beam (of sufficient depth), interconnecting the columns [Pig. 14.191.

............. pedestal

SECTION 'XX'

SECTION 'W'

Fig. 14.19 Beam-slab combined footing

The base slab behaves likc a one-way slab, supported by the beam, and bends transversely under the uniform soil messure actinc!from below. The loads transferred from the slab are resisted by the longitudinal beam. The size of the beam is generally governed bv . .(one-wav) ,. shear at d from the face of the columnJuedestal. For effective load transfer, the width of the footing beam should be made equal to the columnlpedestal width, and it is advantageous to provide a pedestal to the column. The high shear in the beam will usually call for heavy shear reinforcement, usually provided in the form of multi-legged stirrups [Fig. 14.191. The base slab may be tapered (if the span (B - b)l2 is large), for economy. The thickness of the slab should be checked for one-way shear at d (of slab) from the face of the beam. The flexural reinforcement in the slab is designed for the cantilever moment at the face of the beam, and provided at the bottom, as shown in Fig. 14.19. Two-way shear is not a design consideration in beam-slab footings. The top and bottom reinforcement in the beam should conform to the longitudinal bending moment diagram, and development length requirements should be satisfied.

-

-

~

~

699

DESIGN OF FOOTINGS AND RETAINING WALLS

698 REINFORCE0 CONCRETE DESIGN 200

EXAMPLE 14.7

4500

:\

14604

Design a combined footing for two columns CI (400 mm X 400 tm with 4-2 and C2 (500 mm x 500 nun with 4-28 @bars)supporting axial loads P I = 900 P2=1600 IcN respectively (under service dead and live loads). The column CI is afl exterior column whose exterior face is flush with the property. line. The centre-tocentre distance between CI and Cz is 4.5 m. The allowable soil pressure at the base of the footing, 1.5 m below ground level, is 240 W m 2 . Assume steel of grade Fe 415 in columns as well as footing, and concrete of M 30 grade in columns and M 20 grade in footing.

(a)

footing plan

SOLUTION

Footing base dimensions Assuming thc weight of the combined footing plus backfill to constitute 15 percent of the column loads, P, i P 2i AP - (goo+ l6OO)xl.l5 = 11.98 m2 A,??d = 240 4" In order to obtain a uniform soil pressure distribution, the line of action of the I-esultant load must pass through the centroid of the footing. Lct the footing centroid be located at a distance .?from the centre of Cl [refer Fig. 14.20(a)]: Assuming a load factor of 1.5, the factored column loads are: P,,I= 900 X 1.5 = 1350 W, P.2 = 1600 X 1 3 = 2400 M\I P,,, + P,,2 = 3750 kN spacing between columns s = 4500 mm = , s = 2400x4500 = 2880 mm 3750 sl2 = 2250 mm,a ,rcmngulur footing may be provided, with length L = 2(2880 t 200) = 6160 mm Provide L = 6.16 m *width required B ?AIL= 11.9816.16 = 1.95 m Provide B = 2.00 m

\-,

shear force (kN)

cz

(d) bending moment (kNm)

.

Stress resultants in longitudinal direction Treating the footing as a wide beam (B = 2000 mm) in the longitudinal direction the uniformly distributed load (acting upward) is given by q,,B = (P,,I + Pra)IL= 375016.16 = 608.8 W l m [as shown in Fig. 14.20(b)l. The distribution of shear force is shown in Fig. 14.20(c). The critical section for one-way shear is located at a distance d from the (inside) face of CZ,and has a value V,,I= 2400 - 608.8 (1460 + 250 i d ) X 10" = (1359 - 0.6088d) kN The distribution of bending moment is shown in Fig. 14.20(d). The maximum 'positive' moment at the face of column Czis given by M,' = 608.8 x (1.460 - 0.250)'12 = 446 m m The maximum 'negative' moment occurs at the location of zero shear, which is at a distance x from the edge (near CJ of the footing [Fig. 14.20(c)]:

.

-

'.

,'

(e) column stnps as transverse beams ~ i g14.20 . Example 14.7

kNh

SECTION'A-A'

DESIGN OF FOOTINGS AND R h A l N l N G WALLS

(A,,), ,

x = 13501608.8 = 2.2175 m M ,= 608.8 X (2.~175)~/2 - 1350 x (2.2175 - 0.2) = (-) 1227 kNm

*

Thickness of footing based on shear

j

(a) One-way shear (longitudinal): V,,I Assuming z, = 0.48 MPa (for M 20 concrete, assuming p, = 0.50) V,,, = 0.48 x 2000 x d = (9604 N I/;., = V,,, (1359 - 0.6088d) x 10) < 960d (1 2 866 mm

*

(b) Two-way shear The critical section is located dl2 from the periphery of columns Cl and C2 [Fig. 14.20(a)], and the factored soil pressure q,, = (q,, E)lB = 608.812.0) = 304.4 kN1m2. Assuming d = 866 mm, 1350-304.4(0.4+0.866)(0.4+0.866/2)= 1029 kN at column Cl V,"2= 2400-304.4(0.5+ 0.866)~ = 1832 kN at column C2

For square columns, k, = 1.0 a zc2= 1.0 x 0.25$%= 1.118 MPa

1

1 . 1 1 8 ~ ( 1 2 6 6 + 8 3 3 ~ 2 ) ~ ( 8 6=6 )2 8 3 9 x 1 0 ' ~> 1029 kN 1.1 18x(1366X4)x(866) = 5290~10'N>1832kN Hencc, the depth is governed by considerations o i one-way shear alone. Assuming an overall thickness D = 950 mrn and 20 inn1 $ bars with a clear cover of 75 mn1, effective depth d = 950 - 75 - 2012 = 865 mm (very closc to 866 mm required -OK) Check base pressure: Assuming unit weights of 24 !dlm3 for concrete and 18 k ~ l m for ' backfill, gross soil pressure under service loads q = (900 + 1600)1(6.16 x 2.0) + (24 x 0.95) + (18 x 0.55) = 235.6 kNlm2 < q, = 240 kN/m2 -OK. Design of longitudinal flexnral reinforcement Maximum 'negative' moment: M,,-= 1227 kNlm =3

RG

M,,= Ed2

1227x106

= 0.820 MPa

2000x865~

* p , = 0.239 < 0.50' required for one-way shear 'Note: In general, it is not good practice (and often, not economical) lo fix the flexural steel requirement based on shear strength requirements, if the steel requirement is excessive. However, in this sitnation, p, = 0.50 cannot be considered to be excessive.

2

P = ----[I 100 2 x20 4 1 5 - 4 1 - 4.598 x 0298/20]= 0.084 x 10.' (low)

(A,,),,,;,c= 0.0012 ED = 0.0012 x 2000 x 950 = 2280 mm2 Number of 16 mm $ bars required = 22801201. = 12 [Corresponding spacing = (2000 - 75 x 2 - 16)/11 = 167 mm - OK.] :. Provide 12110s 16 n m $ bars a t bottom as indicated in Fig. 14.21. Required development length = 47.0 x 16 = 752 mm,which is available on the side of the column C2 close to the edge of the footing: by placing the bars symmetrically with respect to column C2, the required length will be available on both sides of the section of maximum 'positive' moment.

Limiting two-way shear stress .rC2= kc ( 0 . 2 5 6 )

V," =

, = 0.50 x 2000 x 8651100 = 8650 mm

>(A,,) ,,,,, = 0.001280 Numbw of 20 mm $bars required = 86501314 = 28 [Corresponding spacing = (2000 - 75 x 2 - 20)/27 = 68 mm, which is low but acceptable.] :. Provide 28 nos 20 mm $I bars a t top between the two columns as indicated in Fig. 14.21. Required development length (with M 20 concrete and Fe 415 bars) will he less than L,, = 47.0 x 20 = 940 mm Adequate length is available on both sides of the peak moment section. Maximum 'positive' moment: M,: = 446 kNm (at face of column C2)

=3

1

701

Design of column strips a s transverse beams [Fig. 14.20(e)].

:

(a) Transverse beam under column C1: Factored load per unit length of beam = 135012.0 = 675 kNIm I Projection of beam beyond column face = (2000 - 400)12 = 800 mm : Maximum niomcnt at column face: M,#= 675 x 0.80~12= 216 kNm Effective depth for transverse beam (16 mm $ bars placed above the 16 m m $ : longitudinal bars): d = 950 - 75 - 16 x 1.5 = 851 mm Width of beam = width of column + 0.75d = 400 + 0.75 x 851 = 1038 mm

;

*R= M" = 216x106 = 00.87 MPa (low)

8d2 1038x8512 Provide minimum reinforcement: A,, = 0.0012 bD *A,, = ,0012 x 1038 x 950 = 1183 mm2 Number of 16 mm $ bars required = 11831201 = 6 [Corresponding spacing = (1038 - 75 - 16)15 = 189 mml Alternatively, no. of 12 mm @ bars required = 11831113 = 11 Provide 11 nos 12 mm (I bars

Required development length = 47.0 x 12 = 564

< (800 - 75) lmn available - OK.

There is no need to check one-way transverse shear in this case as the critidal section (located at d = 851 mm from column face) lies outside the footillg, (b) T,ansver:vcbean1 under column C2: Factored load per unit length = 240012.0 = 1200 ~ N I Projection beyond column face = (2000 - 500)/2 = 750 "M Moment at column face = 1200 x 0.'15~/2= 338 wm Width of beam = 500 + 1.5 x 851 = 1777 nun

.

(b) Colunm C2: Limiting bearing stress at i) colomn face = 0.45f,,, = 13.5 MPa (as belorc) ii) footing face = 0.45f,,:

a

[A, = 2 0 0 0 ~, A2 = 500' m m 2 ~

= 0.45 x 20 x 2.0 = 18.0 MPn

~

.

M,, a,?e = 3 3 8 x 1 0 ~ - 0.263 MPa (low) ~d~ 1777x8512 * = 0.0012 x 1777 x 950 = 2026 nllnZ Number of 12 mm $ bars required = 2026/113 = 18 Provide 18 nos 12 ~ n m $bars R e q u i d development length = 47.0 x 12 = 564 1nln is beyolld the column lace. * AS in the previous case, check for one-way shear is not called for. ~ r a n s f e of r force a t column base (a) Colurnn C1: Limiting bearing stress at i) column face = 0 . 4 5 h = 0.45 x 30 = 13.5 MPa

ii) footing face = 0 . 4 5 f , , a [As the column is located at the edge of the footing, A , A2 = 4002 m 2 ] = 0.45 X 20 x 1.0 = 9.0.--.MPa . . ,. < 13.5 MPa Limiting bearing resistance at column-footing interface Fb,= 9.0 X 4002 = 1440 X 103N > P,,, = 1350 W- OK. Hence, full force transfer can be achieved without the need for reinforcemefit across the interface. However, it is desirable to provide some nominal dowels (4 nos 20 mm $0, as shown in Fig. 14.21. ~

= 4.0, l i l ~ t c dto 2.01

> 13.5 MPa

+ F , = 13.5 x500'= 3375 x 10~id\I>P,n=Z4OOirN. 111 this case also, full force transfer can bc achieved without the need for reinforcement across the ititerhce. However, it is desirable to provide some . nominal dowels (4 nos 20 mill $) as shown in Fig. 14.21.

Reinforcement details The mi~iforcemcntdetails are indicated in Fig. 14.21. Some of the longitudi~lal bars at the bottom are shown (arbitrarily) extended across the full length of the footing in order to providc soine nominal reinforcement in the large (otherwise unreinforced) area of concrete between the columns and also to lie np with the also transverse bars under column C1. Nominal transverse reinforcement indicated at top between the columns, in order to tie up with the main long~tudinal bars provided.

is

14.7 TYPES OF RETAINING WALLS AND THEIR BEHAVIOUR As explained in Sectioll 14.1, retaining walls are used to rctain earth (or other material) in a verUcal (or nearly vertical) position at locatio~lswhere an abrupt change in ground level occurs. The wall, therefore, prevents the retained earth from assuming its natural angle of rcposc. This causes the retained earth to exert a lateral pressure on the wall, thereby tending to bend, overlurn and slidc the retaioing wall structuxe. The wall, including its supporting footing, must therefore be suitably designed to be stublc under thc ellects of the lateral earth pressure, and also to satisfy the usual requiremnents of sueilgth and serviceability. Retailling waUs are usually of the followillg types:

1. Gravity Wall [Fig. 14.22(a)1 The 'gravity wall' provides stability by virtue of its own weight, and therefore, is rather massive in size. It is usually built in stone masomy, and occasionally in plain conc~.etc.The tllic!aess of the wall is also governed by the need to eliminate t . or limit the resulting tcnsile strcss to its permissible limit (which is very low In the case of concrete and masonry). Plain concrete gravity walls are not used for bcights exceeding about 3 m, for obvious econonlic rcasons.

' Tile 'middle third rule' is generally applied, wherein fbe wall thickness is inade sufficiently Fig. 14.21 Details of reinforcement, Example 14.7

large, to ensure ihsr the restzltam ihrusf at any cross-section falls within the 'middle third' region of the sectioll

DESIGN OF FOOTINGS

AND RETAINING WALLS

706

2. Cantilever Wall [Fig. 14.22(b)]

(a) gravity wall

(b) cantilever wall

COUNTERFORT

(C)

counterfort wall

(d) buttress wall

F

ABUTMENT

(€4 basement wall

(f)

bridge abutment

Flg. 14.22 Types of retaining wall structures

The 'cantilever wall' is the most common type of retaining structure and is generally economical for heights up to about 8 m. The structure consists of a vertical stem, and a base slab, made up of two distinct regions, viz. a heel slab and a toe slab. All three components behave as one-way cantilever slabs: the 'stem' acts as a vertical cantilever under the lateral earth pressure; the 'heel slab' acts as a (horizontal) caintilever under the action of the weight of the retained earth (minus soil pressure acting upwards from below); and the 'toe slab' also acts as a cantilever under the action of the resulting soil pressure (acting upward). The detailing of reinforcement (on the flexural tension faces) is accordingly as depicted in Fig. 14.22(b). The stability of the wall is maintained essentially by the weight of the earth on the heel slah plus the self weight of the structure. 3. Connterfort Wall [Fig. 14.22(c)] For large heights, in a cantilever retaining wall, the bending moments developed in the stem, heel slab and foe slab become very large and require large thicknespes. The bending moments (and hence stemlslab thicknesses) can be considerably reduced by introducing transverse supports, called counterforfs, spaced at regular intervals of about one-third to one-half of the wall height), interconnecting the stem' with the heel slab. The counterforts are concealed within the retained earth (on the rear side of the wall). Such a retaining wall structure is called the courrterjorr wall, and is economical for heights above (approx.) 7 m. The counterforts subdivide the vertical slah (stem) into rectangular panels and support them on two sides (suspender-style), and themselves behave essentially as vertical cantilever beams of T-section and varying depth. The stem and heel slub panels between the counterforts are now effectively 'fixed' on three sides (free at one edge), and for the stem the predominant direction of bending (and flexural reinforcement) is now horizontal (spanning between counterforts), rather than vertical (as in the cantilever wall). 4. Buttress Wall [Fig. 14.22(d)] The 'buttress wall' is similar to the 'counterfort wall', except that the transverse stem supports, called butfresse.~, are located in the front side, interconnecting the stem with the toe slab (and not with the heel slab, as with counterforts). Although buttresses are structurally more efficient (and more economical) than counterforts, the counterfort wall is generally prefemd to the buttress wall asbit provides free usable space (and better aesthetics) in front of the wall. 5. Other Types of Walls Retaining walls often form part of a bigger structure, in which case their structural behaviour depends on their interaction with the rest of the structure. For example, the exterior walls in the basement of a building [Fig. 14.22(e)] and wall-type bridge abuhncnts [Fig. 14.22(f)] act as retaining walls. In both these situations, 'The toe slab is also frequently interconnected with the stem (in the front side of the wall) by means of a 'front counterfort', whose height is limited by the ground level on the toe side, so that it is concealed and provides free usable space in front of the wall.

706

REINFORCED CONCRETE

DESIGN

the vertical stem is provided an additional horizontal restraint at the top, due to tile slab' at the ground floor level (in the case of the basement wall) and due to the bridge deck (in the case of bridge abutment). The stem is accordi~~gly designed as a beam, fixed at the base and simply supported or partially restrairlcd at the top. The side walls of box culverts also act as retaining walls. In this case, the box culvert (with single/multiple cells) acts as a closed rigid frame, resisting the combined effects o f lateral earth pressures, dead loads (due to self weight and earth above), as well as live loads doe to highway traffic.

D E S OFFOOTINGS ~ A N D RETAINING WALLS

707

of shearing resistance (or artgle of repose). For a Where @ is the granular soil (such as sand), $ = 30U,correspondiag to which, C, = 113 and Ct, = 3SL , as shown in Fig. 14.23, the expressioll [Eq. 14.lOal wlten [he backfill is fnt C. ~houldbe modified as follows:

In the sections to follow, only the cantilever and counteifort retaining walls are discussed - with particular emphasis on the cantilever wall, which is the most common type of retaining wall structure. 14.8 EARTH PRESSURES AND STABILITY REQUIREMENTS 14.8.1 Lateral Earth P r e s s u r e s The lateral force duc to earth pressure constitutes the main force acting on the of the retaining wall, tending to make it bend, slide and overturn. Thc detcrlnit~atiol~ magnitude and direction of the earth pressure is based on the principles of soil mechanics, and the reader may reier to standard texts in this specialised area (such as Ref. 14.2, 14.3, 14.8) for a detailed study. In general, the behaviour of lateral earth pressure is analogou$,to that of a fluid, with the magnitude of the pressure p increasing nearly linearly with increasing depth z for moderate depths below the surface:

P = CY,Z (14.9) where y, is the unit weight of the earth and C is a coefficient that depends on its physical properties, and also on whether the pressure is active or passive. 'Active pressure' (pJ is that which the retained earth exerts on the wall as the emth moves in the same directin11 as the wall deflects. On the other hand, 'passive pressure' (p,)is that which is developed as a resistance when the wall moves and presscs against the carth (as on the toe side of the wall). The coefficient to be nsed in Eq. 14.9 is the active pressure coefficient, C,, in the case of active pressure, and thepassive p,rssu,r coefficient, C,, in the case of passive pressure; the latter (C,)is generally much higher thau the former (CJ for the same type of soil. In the absence of i o r e detailed iofounation, the following e x p ~ ~ s s i o for a s C, and C,,, based on Rankine's thcory [Ref. 14.2, 14.31, may be used for cohesionless soils and level backfills:

I+sin@ c,, =I-sill$

Fig. 14.23 Forces acting on a cantilever retaining wall

~h~ direction of the active pressure, p, [given by Eq. 14.91, is Parallel to the backfill. ~h~ l,rcssurc has a ntaxituu~nvalue at the hcel, and is equal to surface of h'is the lleight of the backEil1. mneasured vertically above the heel cn eh', [pig, 14.231. F~~the case of a level backfill, 8 = 0 and h' = h, and the direction of the lateral pressure is horizontal and normal to the vertical stem. *lte force, p,,, exerted by the active earth pressure, due to a backfill of height h' heel, is accordillgly obtained from the triangular pressure distribution . [Fig. 14.231 8s (14.12) P , ~=cay,(h')'/2 per m ~engtbo f t h c wall, and acts at a height h'/3 .,-his force has units of above the heel at an inclination 0 with the horizontal.

~

' The slab is integrally connected to numerous bean~-colunmframes, and the lateral istraint offered by it is due to the high sror-ey srt@%e.x at the lowenuost s t o ~ y .

-

[refer Section i4.8.21

The force, P,, developed by passive pressure on the toe side of the retaining wall is generally small (due to the small height of earth') and usnally not included in the design calculations, as this is conservative.

where

14.8.2 Effect o f Surcharge o n a Level Backfill Frcqucntly, gravity loads act on a level backfill due to the construction of buildings and the movement of vehiclcs near the top of the retaining wall. These additio~lal loads can be assumed to be statici and uniformly distributed on top of the bac!dill, for calculation purposes. This distributed load bo, (k~lm') can bc treated as statically equivalent to an additional (fictitious) height, it, = w&, of soil bacldill with unit weight y,. This additional hcight of backfill is called srrrclmrge, and is expressed either in terms of height h,, or in terms of the distributed load W , [Fig. 14.24].

~,,=C,w,h=C,%h,h

(14 13a)

Pa2= C. y, h2/2

(14 13b)

wlth the lmes of actlon of P,,, and P, at hl2 and h/3 above the heel 14.8.3 Effect of Water in t h e Backfill

I

:

When water accumulates in the backfill, it can raise the lateral pressure on the wall to very high levels. If the water in the backfill does not have an escape route, it will build up a hydrostatic pressure on the wall, causing it to behave like a dam. The resulting pressuref distributions are depicted in Fig. 14.25.

Fig. 14.25 Effect of water in the backfill

Fig. 14.24 Effectof surcharge on a level backfill The presence of the surcharge not only adds to the gravity loading acting on the heel slab, but also increases the lateral pressure on the wall by C,y,h, = C.w,. The resulting trapezoidal earth pressure distribution is made up of a rectangular pressure distribution (of intensity Cow,), superimposed on the triangular pressure distribution due to the actual backfill, as shown in Fig. 14.24. Thc total force due to active pressure acting on the wall is accordingly given by

'

Strictly, for the full devcloprnent of pnssive earth pressure, it is lnecesrnry that dnring the construction of the wall, there should be no disturbance ta the soil agnhst which the concrete in the toe slab is nlaced. In the case of ~ehiculaitraffic and other live loads, the equivalent loading should include a dynamic magnification factor.

'

14.8.4 Stability R e q u i r e m e n t s The Code (Cl. 20) specifies that the factors of safety against overturning (CI. 20.1) and sliding (CI. 20.2) should not be less than 1.4. Furthermore (as explained in Section 14.3.3), as the stabilising forces are due to dead loads, the Code specifies that

'

The presence of water does not significantly alter the shearing resistance of granular soils: hence the coefficient. C. is practically the same for both dry and submerged conditions.

710 REINF ORC ED CONCRETE DESIGN

these stabilising forces should be factored by a value of 0.9 in calculati~lgthe factor of safety, FS. Accordingly, FS =

0.9 X (stabilising force or ~non~ezlt) 2 1.4 destabilising force or lnoment

Overturning If the retaining wall structure were to overturn, it would do so with the toe acting as the centre of rotation. In an overturning context, there is no upward reaction R acting over the base width L. The expressions for the overturning moment M, and the stabilising (restoring) moment M, depend on the lateral earth press~reand the geometry of the retaining wall. For the case of a sloping backfill [Fig. 14.231,

(FS)rljdi,,g=

0.9F , which should be Z 1.4 P" cos 8

(14.1911)

When active pressures are relatively high (as when surcharge is involved), it will be generally difficult to mobilise the required factor of safety agail~stslidi~~g, by considering frictional resistance bclow the footing alone [Eq. 14.191. In such a situation, it is advantageous to usc a drear key projecti~lgbelow thc lootil~gbase and extending throughout the let~gtllof the wall [Fig. 14.261, When the concrete in the 'shear key' is placed in an unfomed excavation (against undislorhed soil), it can b e expect& to develop considerable passive resistance. Different procedures have been proposed to estimate this passive resistance P;, [Ref. 14.8, 14.91. A simple and conservative estimate is obtained by considering the pressure developed over a region, h,- h,, below the toe: (14.20) !

where h, and h2 are as indicated in Fig. 14.26. It may he noted that the overburden due to the top 0.3 rn of earth below ground level is usually ignored in the calculation.' where W denotes the total weight of the reinforced eoncmte wall structure plus the retained earth resting on the footing' (heel slab), and x,, is the distallce of its lirle of action from theheel, as shown in Fie. 14 2 1 ~ ". For the case of a level backfill with surcharge [[Fig. 14.241,

300 mm overburden

M, = P,1(h/2) + P,,(h/3) (14.17) where Po,and Pa, are as given by Eq. 14.13(a) and Eq. 14.13(b) respectively. The expression for M,is the same as that given by Eq. 14.16, but with Q = 0. The factor of safety required against overturning [Eq. 14.141is obtained as (Fs)ovmnd,u =

O.9Mr -2 1.4 M,

(14.18)

Sliding The resistance against sliding is essentially provided by the friction between the base slab and the supporting soil, given by

F=pR (14.19) where R = W is the resultant soil pressure acting on the footing base and p is the coefficient of static friction between concrete and soil. [In a sloping backfill, R will also include the vertical component of earth pressure, P, sin0 (see Fig. 1 4 . 2 3 ) ~The value of p varies between about 0.35 (for silt) to about 0.60 (for rough rock) [Ref 14.21. The factor of safety against sliding [Eq. 14.141 is obtained as 'The weight of the d h f i l l above the toe slab is usually (conservatively)ignored. Similarly,

the passive eefh pressure P, is also usually ignored.

Fig. 14.26 Passive resistance due to shear key I'

The shear key is best positlonetl at n distance xSkfrom the toe in such a way that the flexural reinforcement fiom thc stem can be extended straigllt into the shear key near the toe. 14.8.5 Soil Bearing Pressure Requirements

The width L of the base slab nust be adcquate to distribute the vertical reaction R to the foundation soil without causing excessive settlenlent or rotation. As explailled in Section 14.3, thc required foundiug depth and the associated allowable pressure qy,are usually prescribed by a gcotcchnical consultant on the basis of a soil study, and the control on vertical settlelnent is built into thesc rccommcndations. Ilowever, the designer must further a m r e that tilting of the f o o t i ~ ~isgalso avoided by avoiding a highly non-uniform base pressure in weak soils.

DESIGN OFFOOTINGS AND

712 REINFORCED CONCRETE DESIGN 14.9 PROPORTIONING AND DESIGN OF CANTILEVER AND COUNTERFORT WALLS

Prior to carrying out a dctailcd analysis and design of the retai~~ing wall structure, it is nccessary to assume preliminary dinlensions of the various clcmcnts of thc structure using certain approximations. Subsequcntly, these dimensions may bc suitably revised, if so required by design considerations.

WALLS 713

will be uniform if L is so selected as to make aR= 0.5. Similarly, for c& = 213, the base pressure distribotion will be triangular. Thus, fol: any selected distribution of base pressure, aRis a constant and the required base width L = LRlaR. Considcring static equilibrium and taking moments about reaction point e, and assuming X,, = a d 2 ,

y, h ~ ((a, ' a, - a 3 2 ) = C,& h3/6

14.9.1 Position of Stem on Base Slab for Economical Design An important consideration in thc dcsign of cantilever and counterfort walls is the position of the vertical stem on the base slab. It can be shown [Ref. 14.101 that an economical design of the retaining wall can be obtained by proportioning the base slab so as to align the vcrtical soil reaction R at the base with thc front face of the wall (stem). For this derivation, let us consider the typical case of a level bacW11l [Wg. 14.271. The location of thc resultant soil reaction, R, is dependent on the magnitude and location of the resultant vertical load, W, which in turn depends on the dimension X (i.e., the length of heel slab, inclusive of the stem thickness). For convenience in the derivation, X may be expressed as a rraction, a,, 01the lull width L of the base slab (X = a,L). Assuming an avcragc unit wcighr y, lor all matcrial (earth plus concretc) behind rhc front face of the stem (rcctangle obcd), and neglecting entirely the weight ofcuncrcte in the toc slab, R = W=y,hX=y,h(axL)

RETAINING

For economical proportioning for a given height of wall (h), the length of the base (L) must be minimum, i.e., Uh should be minimum. From Eq. 14.21, this implies that (2 a" ax - a 3 should be maximum. The location of R, and hence thc base width for any selected pressure distribution, is dependent on the variable X, i.c., ax. For maximising (2an ar- $J, ax= a~ *a,L

= cl,L=X

Width of Base Applying the above principle, an approximate expression for the minimum length of base slab for a given height of wall is obtained from Eq. 14.21 as:

Alternatively, thc minimum width of heel slab is given by:

Fig. 14.27 Proportioning of retaining wall For a given location of R col~espondingto a chosen value of X, the toe projection of the base slab (and hencc its total width, L) can be so selectctl by the designer qs to give any desired distribution of base soil pressure. Thus, representing the distance, L,, from the heel to R as a lraction a, of base width L, [Fig. 14.271, the base pressure

The effect of surcharge or sloping backfill may be taken into account, approximately, by replacing h with h + h,, or It', respectively. Alternatively, and perhaps more convenienlly, using the above principle, tile heel slab width (X in Fig. 14.27) may be obtained by equating moments of Wand Pa about the point (1. The required L can then be worked out based on the base pressure distribution dcsired. It may bc noted that thc total height h of the retaining wall is the dilfercnce in elevation between thc top of thc wall and the bottom of base slab. The lattcr is based

714 REINFORCED CONCRETE DESIGN

DESIGN OFFOOTINGS AND on geotechnical considerations (availability of firm soil) and is usually not less than I m below the ground level on the toe side of the wall. After fixing up the trial width of the heel slab ( = X) for a given height of wall and backfill conditions, the dimension L may be fixed up. Initially, a triangular pressure 3 distribution may be assumed, resulting in L = - X . Using other approximations

-

2

(discussed in the next section) related lo stem thickness and base slab thickness, a proper at~al~sis* should be done to ascertain that (1) the factor of safety against overturning is adequate; (2) the allowable soil pressure, q,, is not exceeded; and (3) the factor of safety against sliding is adequate. Condition (1) is generally satisfied; however, if it is not, the dimensions L and X may he suitably increased. If condition (2) is not satisfied, i.e., if q,,,, > q,, the length L shotlld be inc~easedby suitably extending the length of the toe slab; the dimension X need not be changed. If condition (3) is not satisfied, which is usually the case, a suitable 'shear key' should he designed. .,..:,

14.9.2 Proportlonlng a n d Deslgn of E l e m e n t s of Cantllever Walls

+

: :

Initial T h i c k n e s s of B a s e Slab and S t e m For preliminary calculations, the thickness of the base slab may he taken as about 8 percentof the height of the wall plus surcharge (if any); it should not be less than 300 m m The base thickness of the vcrlical stem may be taken as slightly more than that of the hase slab. For economy, the thickness may be tapered linearly to a minimum value (but not less than 150 inm) at the top of the wall; the front face of the stcm is maintained vertical'. If the length of the heel slab and/or toe slab is excessive, it will be economical to provide a tapered slab. With thc above preliminary proportions, the stability check and determination of soil pressure (at the base) may be performed, and ditner~sionsLand X of the base slab [Fig. 14.271 finalised. It may be noted that changes in thicknesses of base slab and stem, if required at the design stage, will he marginal and will not affect significantly either the stability analysis or the calculated (gross) soil pressures below the hase slab. Design of Stem, Toe Slab and H e e l Slab The threc elements of the retaitling wall, vie., srcnr, toe slab and heel slab have to be designed as cantilever slabs to resist the factored moments and shear forces. For this a load factor of 1.5 is to be used.

'

In such an analysis. it will be seen that the zcmal vertical rcaction R below the foatinp base will he close to, although rarely coincident with, thc front face of the stem (as assumcd initially). It is recommended that a batta of 1 : 50 be provided to the front face of the stem during onstniction. to offset the deflection of the stem or possible forward tilting of the structure [Ref. 14.101.

RETAINING WALLS

715

In the case of the toc slab, the net pressure is obtained by deducting the weight8 of the concrete in the toe slab from the upwad acting gross soil pressure. The net loading acts upward (as in the case of usual footings) and the flexural reinforcement has to be provided at the bottom of the toe slab. The critical section for moment is at the front face of the slem, while t l ~ ecritical section for shear is at a distanced from the face of thc stem. A clear cover of 75 mm may be provided.in base slabs. In the case of the heel slab, the pressures acting downward, due to thc weight of the retained earth (plus surcharge, if any), as well as the concrete in the heel slab. exceed the gross soil pressures acting upward. Hence, the net loedittg acts downward, and the flexural reinforcement has to be provided at the top of the heel slab. The critical section for moment is at the rear face of the stem base.

In the case of the stem (vertical cantilever), the critical section for shear lnay be taken d from the face of llic support (top of base slab), while the critical section for moment should be taken at the face of the support. For the main bars in the stcm, a clear cover of 50 nun may be provided. Usually, shear is not a critical design consideration in the stem (unlike the base slab). The flexural reinforcement is provided near the rear face of the stem, and may bc curtailed in stages for economy [refer Example 14.91 Temperature and sltrinkage rcinforcenlent (A,,,,,,fl,,= 0.12 pcrcent of gross area) should be providcd tramvcrse to the main reinf~rccment. Nominal vertical and horizontal reinforcement should also be provided near the front face which is exposed.

14.9.3 Proportioning a n d Design of E l e m e n t s of a Counterfort Wall Initial T h i c k n e s s e s of Various Elements In a counterfort wall, counte~fortsare usually pmvided at a spacing of about one-third to one-half of the height of thc wall. The triangular shaped counterforts are provided it, the rear side of the wall, interconnecting the stem wit11 the heel slab. Sometimes, small buttresses are provided in the front side below the ground level, interconnecting the toe slab with the lower portion of the stem. The presence of countcrlorts enables the use of stem and base slab thicknesses that are much smallw. tllan lhose normally required for a cantilever wall. For preliminary calculations, the stcnl thickness and hcel slab thickness may be taken as about 5 percent of l l ~ cheight of the wall, but not less than 300 m m If the front buttress is provided, the thickness al' thc toe slab may also be taken as 0.05h: olhcrwisc, it may bc taken as in the case of lhc cantilever wall (0.08h). Thc Lhickncss of the

'

' R e weight of the earthfill io this legion is (conservatively) ignored

716 REINFORCED

CONCRETE DESIGN

counterforts may be taken as about 6 perceot of the height of the wall at the base, but not less than 300 mm. The thickness may be reduced along the height of the wall. With the above preliminary proportions, the stability check and determination of soil pressures (at the base) may be performcd, and dimensions L a n d X of the base finalised, as in the case of the cantilever wall.

Design of Stem, Toe Slab and Heel Slab Each panel of the stem and heel slab, between two adjacent counterforts, may be designed as two-way slabs fixed on three sides, and free on the fourth side (free edge). Thcsc boundary conditions are also applicable to the toe slab, if buttresses are provided; otherwise the toe slab behaves as a horizontal cantilever, as in the case of the cantilever wall. The loads acting on these elements arc identical to those acting on the cantilever wall discussed earlier. For the stcm, bending in the horizontal direction between counterforts' is gcncrally more predominant than bending in the vertical direction. Near thc counterforts, the main rcinforccment will be located close to the rear face of the stem, whereas midway between counterforts, the reinforcement will be close to the outside face; the latter is indicated in Fig. 14,22(c). These two-ways slabs, subject to tria~~gt~la~~ltrapezoidal pressure distributions may be designed by the use of moment and shear coefficients (based on plate theory), available in various handbooks, and also in the IS Code for the design of liquid storage structures, viz., IS 3370 (Part 4) [Ref. 14.1 11. Alternatively, the slabs may be designed by the yield line theory. An alterantive silnplificd method of analysis is demousu.ated in Example 14.10.

Design of Counterforts The main counterforts should be firmly secured (by additional ties) to the heel slab, as well as to the vcrtical stem, as the loading applied on these two ele~nentstend to scparate them from the counterforts. 111additionthe counterfort should be designed to resist the lateral (horizontal) force transmitted by the stem tributary to it. The counterfort is designed as a vertical cantilever, fixed at its base. As the stem acts integrally with the counterfort, the effective section resisting the cimtilever moment is a flanged section, with the flange under con~p~ession. Hence, the counterforts may be designed as T-beams [refel. Chapter 51 with the depth of section varying (linearly) from the top (free edge) to the bottom (fixed edge), and with the main reinforcement provided close to the sloping face. Since these bars are inclined (not parallel to the compresssioll face), allowance has to be made for this in computing the area of steel required.

' An

~pproxilnateand conservative estimate of this bending mamnent can be obtained by treating the slab as one-way continuous slab spanning the eounterfons.

EXAMPLE 14.8

Determine suitable dimensions of a cantilever retaining wall, which is required to support a bank of earth 4.0 m high above thc ground level on the toe sidc of the wall. Consider the backfill surfacc to bc incli~~ed at an angle of 15' with thc horizontal. Assume good soil for foundation at a depth of 1.25 m below the ground level with a safe bearing capacity of 160 !di/m2. Further assume the backfill to comprise granular soil with a unit weight of 16 k ~ / and ~ n an~ angle of shearing resistance of 30'. Assume the coefficietlt of friction between soil and concrete to be 0.5. SOLUTION

1. Data given:

.

. . .

h = 4.0 + 1.25 = 5.25 I"; p = 0.5 8 = 15" y, = 16 kNlm3

Earth pressure coefficients: C, =

- JcosA e - COS' @

cosB = 0.373

l+sinO C,, =---- = 3.0 1-sin 0

2. Preliminary proportloi~s Thickness of footing base slab = 0.08h = 0.08 x 5.25 = 0.42 m Assume a thickness of 420 mm. Assume a stem thickness of 450 mm at the base of the stcm, lapering to a value of 150 mm at the top of the wall. For an economical proportioning of the length L of the base slab, it will be assumed that the vertical reaction R at the footing base is in line with the Cront face of the stcm. For ~ u c ha condilion, (assuming the height above top of wall to be about 0.4 m), the length of tbc heel slab (inclusive of stem thickness) [Eq. 14.231: (5.25 + 0.4) = 2.0 m X= Id =

(m)

Assuming a triangular basc pressure distribution, L=1.5X=3.0m The preliminary proportions are shown in Fig. 14.28(a). Stability against overt~iri~ing Force due to active prcssore: Po = C,y, h"/2 where h' = h + XtanO [Fig. 14.28(a)1 = 5250 + 2000 tan IS0= 5786 nun Pa = (0.373)(16)(5.786)~/2= 99.9 kN (per in lc~lgthof wall) s P, cos B = 99.9 cos 15" = 96.5 kN Pasin 0 = 99.9 sin 15" = 25.9 kN Overturt~ingmoment Mu= (P,< cos8)h1/3 = (96.5)(5.78613) = 186.1 !dim Line of action of resultant of vertical forces [Fig. 14.28(a)] with respect to the heel can be located by applying statics, considering l m lc11gtI1of tlle wall:

DESIGN 718 REINFORCED

OF FOOTINGS AND RETAINING WALLS

719

CONCRETE DESIGN

.. .,

4. soil pressures a t footing base [refer Rg. 14.28@)1

(a) forces on wall

(with preliminaly

proportions)

and

,,

-

= 232.9 (1 - 0.578) = 32.8 k ~ l m '[refer Fig. 14.28(b)1

3.0 5. Stability against sliding Sliding force = P, c o d = 96.5 kN ~esistingforce (ignoring passive pressure on the toe sidc) F = W = 0.5 x 232.9 = 116.4 W

..

. . 300 neglect

resultant vertical reaction R = W = 232.9 kN (perm length of wall) distance of R from heel: L, = (M,, + M , ) / R = (230.6 + 186.1)/232.9 = 1.789 mt cccenvicity = L, - u 2 = 1.789- 3.012 = 0.289 m, < Ll6 = 0.5 H ~the restlitant ~ ~ lies within ~ , the middle third of the base, which is desirable

-0----

be provided to mobilise the balance force though passive H ~a ~ key~may ~ , resistance. A~~~~~~a shear key 300 mm x 300 mm, at a distance of 1300 mm from toe as shown in ~ i14.28(c). ~ . Distance hz = 0.950 +z300 + 1.300 tan 30°= 2.001 m P,= cny,(hZ2- hZ1)12 = 3 x 16 x (2.001' - 0.95 )I2

EXAMPLE 14.9

(b) calculation of soil pressures

Fig. 14.28 Example 14.8

(c) design of shear key

Repeat the problem in Example 14.8, considering the backfill to be level, but subject to a surcharge pressure of 40 kN/m2 (due to the construction of a building). Design the retaining wall structure, assuming M 20 and Fe 415 steel. SOLUTION

1. Data given: (as in Example 14.8) =1distance

0

of resultant vcrtical force from heel X,y = M ~ I W = 230.61232.9 = 0.990 m Stabilising moment (about toe): M,= W(L-qv) = 232.9 X (3.0 - 0.99)

Note that this value of Ln is different from, although close to, the value of X = 2.0 n1 asssumcd in the initial proportioning.

720 REINFORCED

CO NC RE TE

DESIGN

a Equivalent height of earth as surchavge, h, = forces on wall (wllh prellm~nary proponlons)

1"

Y,

=

40 = 2.5 lm 16

I- sing Earth pressure cocflicients: C, = 7 = 113 I +Sill$ C,=l/C, =3.0 Preliminary proportions Thickness of footing base slab = 0.08 (h + h,) = 0.08 x 7.75 = 0.620. Assume a thichiess of 620 nun. Assume a stem thickness oI 650 nnn at the base of the stem, tapering to a value of 200 rmn at the top of the wall. For an economical proportioning of the length L of the base slab, it will be assutnerl that the vcnical reaction R at the footing basc is in line with the front face of the stem. For such a condition, the lcngth of the heel slab (inclusive of Stem thickness) X =m ( d +hs) = @ , @(77.5) = 2.58 m

45.7kNIm2

(b) calculation of

soil pressures

300 neglected (c) design of shear

key

Lct X = 2.6 m. Assuming a triangular soil prcssurc distribution below the base, L= 1SX= 1.5x2.6=3.9m The preiitninary proportiom are sltown in Fig. 14.29(a) Stability against overturning Forces due to activc pressure (pcr in length of wall) [Fig. 14.29(a)1: POI= C, w, h = (1/3)(40)(5.25) = 70.0 kN POz= C, y, 212 = (1/3)(16)(5.25)~/2= 73.5 id'J =,Pa = 70.0 + 73.5 = 143.5 kN Overturning moment M, = P,,,1112+ Pn h13 =, M, = (70.0)(5.25/2) + (73.5)(5.25/3) = 312.4 !&in (per m length of wall) Line of action of resultant oI vertical forces [Fig. 14.29(a)] with respect to thc heel cat1 be located by applying statics, consideritlg 1 111length of the wall:

722

REINFORCED CONCRETE

DESIGN

3 distance

oi resultant vertical force from heel xlv = MIV/W=525.51366.8= 1.432 rn Reierring to Fig. 14.29(b), r Stabilising moment (about toe): M, = w (L - xlv) = 366.8 X (3.9 - 1.432) = 905.3 kh'm (pcr m length of wall) 0.9M 2 0 . 9905.3 ~ (FS)oper~~,n,a, = 2.61 > 1.40 - OK Ma 312.4 4. Soil pressures a t footing base [refer Fig. 14.29(b)1 resultant ve~ticalreaction R = W = 366.8 kN (perm length of wall) distance of R from heel: L, =(MI, + M,)/R = (525.5 + 312.4)/366.8 = 2.284 m' e eccentricity e = L,*- LIZ = 2.284 - 3.912 = 0.334 m (< U6 = 0.65) indicating that the resultant lies wcll inside the middle third of tllc base

-

DESIGN

OF FOOTINGS AND RETAINING WALLS

6. Design of toe slab The loads considered for thc design of the toe slab are as shown in Fig. 14.30(a). The nct pmsures, acting upward, are ubtaincd by reducing the unifornlly distributed self-weight of thc toe slab from the gross pressures at lhc base. Self-weight loading = 25 x 0.62 = 15.5 kN/m2 The net upward pressure varies from 126.9 kN/mn2 to 94.7 kN1mz, as shown in Fig. 14.30(b). Assuming a clear cover 0175 nun and 16 $bars, (1 = 620 - 75 - 8 = 537 mm Applying a load factor of 1.5, tltc design shcar Corce (at d = 537 nun Crom tlte front face of the stem) and the desigu moment at tlte face of the stem are give11 by: V,,=1.5(126.9 + 94.7)/2 x (1.3 - 0.537) = 126.8 liN/m M,, = 1.5 x r(94.7 x 1.3~12)+ (126.9 - 94.7) x 0.5 x 1.3' x 2/31 = 147.2 kNm/m

For a z, = 0.24 MPa, thc required p, = 0.1 0 wit11 M 20 conclete [reCer Eq. 6.11

as shown in Fig. 14.29(b). 5. Stability against sliding r Sliding force = P,, = 143.5 kN (pcr m length of wall) o Resisting force (ignoring passive prcssure) I.'= @ = 0.5 X 366.8 = 183.4 liN > P, 0.9F 0.9x183.4=1,15ning betwecn the counterforts. The bending nlonrents reduce along the height of the stem, owillg 10 the reduction in the lateral pressures with increasi~lgheight.

b

(A,,),*,,,, = (0.173 x lo-') x lo3 X 417 = 721 mm2/m > (A,,),,,,, = 600 imn2/m Spacing of 12 4bars required = 113 x 10'/72l = 156 mm Provide 12 $ bars @ 150 d c at the bottom of the heel slab throughout. Distribution stecl:

*

Provide 12 $ bars @ 180 c/c at the top of the heel slab throughout.

Design of heel slab for cantilever action Consider the triangular loading on the heel slab [Fig. 14.34(a)1 to be catried by cantilever action with fixity at the face of the stem. The intensity of load at the face of the stem = 47.3 kN/m2. The intensity of load at a distance of 1.5m from the face of the stem is 85.9 kN/m2. Total B.M.due to loadina on the trianeular oortion

.

Intensity of earth pressure at thc base of the stem is p, = C, y, h = (0.333)(16)(8.5) = 45.33 k ~ l n ?(linearly varying to zero at the top) Applying a load factor a i 1.5, w,, = 1.5 x 45.33 = 68.0 k ~ l m at ' base Clear spacing betwcen the countcrforts = 3.0 m. Design of stem for continuous beam action At base Assuming a clcar cover of 50 111111 and 20 $bars, d = 600 - 50- 10 = 540 inn^ and effective span, 1 = 3.0

~

This moment is distributed non-ulliformly across the width of 3.0nl. For design purposes, the max. moment intensity (in the middle region) may be taken as two times the average value

+ 0.54 = 3.54 111

Max. -ve moment occurring in the stem at the counterfort location is given by M,,,.,, = w,,12/12= 6 8 0 x 3.542/12= 71.0 kNln/lll Max, mid-span niomenlnmy bc taken as M..,,.A,.. . .. = w,.12/16= 0.75 X M,,., = 53.3 kNd1n ..-. Design shear force V,,= w,, x (clearspan12 - (1) = 68.0 x (3.012 - 0.54) = 65.3 kN11ll ~of

(rear ~ face), reinforcement ~ ~ for 1 -ve moments at the collnterf~rts

736

REINFORCED CONCRETE

0.12 Min. A,, = -(1000~600) 100 Check for shear at basc

DESIGN OFFOOTINGS

DESIGN

AND RETAINING WALLS

737

= 720 mm21m > 369 mm2/m

T, = 65'3x103 = 0.121 MPa < T, = 0.29 MPa (for mini~mt~np, = 0.15) - OK

10~x540 (Evidently, it is possible to reduce the thickness of the stcm, for economy).

Desien of (front face) reinforcement for +ve moment's in the mid-soan of stem The minimum reinforcement requirement will goverti the design on both faces, < IM,,,,.,. since M,,,,, Using 12 $ban, spacing required = 113 x 10001 720 = 156 mm Provide 12 t$ bars (horizontal) 8 150 clc on both faces of the stem (up to onethird height above base). At ose-tltbd heigllt nbove bose

,

-

d = 500 - 50 - 6 = 444 lnm and effective span I = 3.444m A t,,,, =n~,,1'112=(68.0~213)~(3.444)~/12=44.81 kNmlm

Fig. 14.35 Loading considerations for simplified analysis of stem The intensity of horizontal pressure a[ the base of the stem = 45.3 liN/rn2. The intensity of l~orizontalpressuw at a distance of 1.5 m fro111 the base of the stem is 37.3 k ~ l m ' . 0.12 Min. A,, = -(1000X500) = 600 mm2/m> 282 mmzlm 100 Using 12 $ bars, spacing reqtlired = 113 x 10001 600 = 188 rnm Providc 12 $ hars (horizontal) @ 180 clc on hot11 faccs of the stem (in the nidNe one-third height). At two-thlrir~lsheight abose bnae

0.12 Min. A,, = - (1000X400) = 480 mm72/m 100 Using 10 $bars. spacing required = 78.5 x 10001 480 = 163 nlm Using 12 $bars, spacing required = 113 x 10001 480 = 215 nun Provide 12 r$ hars (horizontal) 8 230 clc on both faccs or the stem (in thc llpper one-third height).

Total B.M. due to loading on the triangular portion = 67.5 kNln 2x3 This nlolnent is distributed non-uniformly across the width 013.0n1. For design purposes, the max. moment intcnsity (in the middle region) may bc taken as lwo times tile average value =, M,>,,=2x(67.513.0) = 45.0 !dindm

effective depth d = 515 - 12 = 503 mm

( I J Z )~ ~25~ -~ --b-41-4.598x0.267/25] 100

e

Design of stem for cantilever action Considcr the triangular loading on thc stem [Fig. 14.351 lo bc carried by cantilever action about the lace of thc stem as follows:

,

2x415

.... ~ , .

temperature and shrinkage)

r: 0.075

x 10" (requited up to

DESIGN

738 REINFORCED CONCRETE DESIGN

The mini~nulnreinforcement requirement will govern the design. Provide 12 $ bars (vertical) @ 150 c/c on both faces of the stem tljl018gh oel ihe height of the stem. Thc reinforcelnent details for the stem, toc slab and heel slab are shown in Ag. 14.36

OF FOOTINGS AND RETAINING

WALLS 739

Clear spacing of counterforts = 3.0 m Thus, each counterfort receives earth pressure from a width of I= 3.0 + 0.5 = 3.5 m At base The intensily of earth pressure at the base of the Stem is p, = C, y, h = (0.333)(16)(8.5) = 45.33 m / n ? Applying a load factor 011.5,

Fro~nFig.14.37. tan 8 = 2700/8500 a 8 = 17 6' and Dh, = 2400 x cos8 = 2287 lm Assuming a clear cowl of 50 111111and 25 $bars, d = 2287 - 50 - 12.5 = 2224 nun

12 $ @ 180clc + addl 5 bars in l m span from free edge

Fig. 14.36 Reinforcement details of stem, toe slab and heel slab 9. Design of interior coonterfort Thc typical interior counterfort acts as a T beam of varying section cantilevel-ing out of thc base slab. The design should include: provisioll for beam actiou provision of horizontal ties against separation frotn stem provision of vertical ties against separation of base o Design of connterfurt for T.benm action

The thickness of countcrforts = 500 mm Clear spacing of countcrforts = 3.0 m Thus, each counterlort receives earth pressure from a width of

Fig. 14.37 Depth consideration for analysis of ~ountelfOrt

DESIGN

740 REINFORCED CONCRETE DESIGN

Effectweflange width (C123.1 2 Code): bf = lo 16+b,, t 6 D f [Bq 4.301

= 850016+500+(6x600) =5517 mm, bf = b ,+ clear span of slab =500+3000=3500mm Thus, b y = 3500 mm (least of the above two values) Approximate requirement of tension steel is given by assuming a lever armz to be the larger of 0.9d = 20011nmand d - D,l2 = 1924mn1, i.e., 2001inin:

3967 No. of 25 $ b a s ~equned= - = 8 bars (provxie in two layers, wlth 25 $ 491 spacer hars) =,d = 2287 - 50 - 25 - 12.5 = 2199 mm Assuming the neutral axis to be located at x,, = D,, MilR= 0 . 3 6 2 ~ 2 5 ~ 3 5 0 0 ~ 6x0(21990 0.416~600)= 37048 x lo6 Nfnm > M,, = 2866 x lo6 Nmm This clearly indicates that the neutral axis lies within thc flange. M ' = 2866x10~ = 0.169 MPa bd 2 3500x2199~

*

(A,),,, = (0.047 x lo-') x 3500 x 2199 = 3639 1iu1?11n (which is close to the aooroximate value of 3967 mm2calculated) A, - 0.85 Minimum reinforcen~entin a beam is given by - -bd f,,

..

Provide 8 nos 25 $ bars in two layers, four bars in each layer with a 25 mm separation.

Above one-llrird lregltt from the base The intensity of earth pressure at h (= 8.5 x 2 13) = 5.67m fioln top is 2 pa = C. h = 45.33 x 2 1 3 = 30.22 kNIm Applying a load factor of 1.5,

V,, = 1.5 x

OF FOOTINQS AND

RETAINING WALLS 741

Assuming a clear cover of 50 mm and 25 $ bars, d = 1525 - 50- 12.5 = 1462 mm Approximate requirement of tension steel is given by assulning a lever arm z to be the larger of 0.9d = 1316mm and d - Df/2= 1212 mm, i s . , 1316 m m :

1789 No. of 25 $ bars required = 49 1

-

4 bars

Assuming the neutral axis to be located at x,, = D,, M,m = 0.362X25X3500~500X (1462 - 0.416X500) = 19860 X 10' N>~.=850xl0~~mm This clearly indicates that the neutral axis lies within the flanee.

-

- - [L-

(P,),~(I 25 ,/I-4.598x0.114/25] = 0.032 x 10-2 100 2x415 (A,,),,, = (0.032 x x 3500 x 1462 = 1638 mm21m (which is close to the approximate value of 1789 imn2calculated) A* .-- 0.85 Minimum reinforcement in a beam is given by ==)--

-

bd

f,

Curtail 4 nos 25 $bars and extend 4 nos 25 $bars (rear face). In order to satisfy the minimum reinforcement criteria, 4 nos 25 $ bars may be extended to the top of the counterfort, without any further curtailment. Design of horizontal ties Horizontal tie (closed stinup) reinforcement in the counterfort serves as shear reinforcement against flexural shear in the counterfort and also as ties resisting the separation of the stem from the counterfort due to the lateral pressure.

At base Shear reinforcement requirement: M V,,,,,,, =V,,-&tan0 d

VUt w = 5 9 8 x 1 0 ~ Nominal shear stress r, == 00.544MPa bd 500x2199

742 REINFORCED CONCRETE DESIGN DESIGN OFFOOTINGS AND RETAINING

*

= I00 A,,lb = 100 x (8x491) l(500x2199) = 0.357 r, = 0.416 MPa

Hence, shear reinforcement is to be provided for a shear force of V,,,=(z,-rJbd= (0.544-0.416) 500x2199 = 140.8 x 10'N Assuming 10 $ 2-legged stirrups. A,,, = 2 x 78.5 = 157 mm2 Required spacing = s, =

0 . 8 7 f ~ A ~-~ d0.87x415x157x2199 = 885

-

l40.8x1o3 0.75d Max, spacing specified by Code = = 3001m (lesser value) 300 mm Tic connection requirement: The tcnsion resisted by the tie reinforcement is given by the lateral pressure on tile wall multiplied by the tributary area. At the base = 45.33 M\T/I~'), the tensile force intensity is accordingly given by: T = 45.33 !&/m2x 3.5m = 158.7 kN/m Applying a load factor of 1.5, the total area of reinforcement required to resist this 1.5~158.7~10' direct tension = = 660 nun2/m. 0.87~415 Spacing of 10 $ 2 legged stirrups required = 2 x 78.5 x lo3/ 660 = 237 mm This tie reinforcement requirement governs (compared to shear reinforcement requirement). VLts

Spacing of 10 $ 2 legged ties required = 2 x 78.5 x 10' / 1400 = 112 111111 Provide 1 0 9 2 legged vertical ties @ 100 null clc up to l m from the free edge. The spacing may be increased to 1.50 mm beyond 1111, owing to the significant reduction in nct pressure. The counterfort reinforccmetlt details ale shown ill Fig. 14.38 and Fig.14.39,

toe slab

Provide 10 $ 2 legged stirrups @ 200 mm c/c in lower one-third region The tie reinforcement requirement will vary linearly along the height of the stem, as the lateral pressure variation is linear. Af one-third height frorrr the base (A,,),,, = (213) x 660 = 440 mm2/m Spacing of 8 @ 2 legged stirrups required = 2 x 50.3 x lo31440 = 228 nl Provide 8 $ 2 legged stirrups @ 200 nun c/c above one-third height. e

WALLS 7 4 3

fi

Design of vertical tics As in the case of the connection between the counterfort and the vertical stem, the connection between the counterfort and the hecl slab must be designed to resist the tension arising out of the net downward pressures acting on the heel slab [Fig. 14.34bl. Considering a l m strip from the free edge, the average downward pressure is (83.4 + 109.1)/2 = 96.25 k~lmn', and hence the average tensile force intensity is: T = 96.25 W11n'x 3.5m = 336.9 W / m Applying a load factor of 1.5, the total area oireinforcement required to resist this 1 . 5 ~ 3 3 6 . 9 ~ 1=0 1400 ~ m2,m direct tension = 0.87x415

~ 2 0 0 0 _ _ f . I

600

I

7

2

4

~

-

I

Fig. 14.38 Reinforcement details of stem and counterfort

DESIGN OF FOOTINGS AND RETAINING

744 REINFORCED CONCRETE DESIGN

W A L L S 745

REVIEW QUESTIONS 14.1 14.2

8 0 2 legged horizontal

10 4 2 legged horizontal stirru~s@ 200 c

separators @ Im

U Fig. 14.39 Section through counteriort showing counterfort reinforcement

What are the main requirements of a foundation system for a structure? Why is it necessary to ensure, by proper proportioning of footings, that the bearing pressures underlying all the footings in a building are more-or-less of the same older of magnitude? 14.3 What arc the situations in which conr6irir.dfootings are preferred to isolated footings? 14.4 Distinguish among the terms (i) allowuble soil presstrre (ii) grass soil pressure (iii) net soil pressure, (i;) factorer( soil pressure. 14.5 What is meant by eccentric loading on a footing, and umider what circumstances does this occur? 14.6 Why is it desirable to eliminate eccentricity in loading on a footing, wherever possible, by means of proper proportioning? 14.7 From structural analyses, it is found that the following stress rcsultants develop at a column base undn the action of chnrucrerisfic loads: (i) P = 475 kN, M = 35 k N p under dead loads; (ii) P = 380 !&M = 39 kNm under live loads; (iii) H = f 30 kN, P = f12 kN, M = ? 41 kNm under wind loads. Determine the combined loads to be considered in deciding the area of the footing to be located in a soil with an allowable soil pressure of 200 WUm2 at a depth of 1.5 m below ground level. 14.8 What arc the advantages of providing pedestals to columns? 14.9 Briefly explain the conditions in which transfer of forces at the interface of colnmn (or pedestal) and footing can be achieved without the aid of reinforcement. 14.10 Under what circumsiances is a trapezoidal shape preferred to a rectangular shape for a two-column combined footing? 14.11 Describe briefly t h e load transfer mechmism in a two-column combined footing. 14.12 What is the purpose of a rztuining wall? What are the different types of concrete retaining walls? 14.13 Distinguish between active pressure and pnssive pressure of earth, in relation to retaining wall structures? ,.14.14 What is meant by (a) surcharge (b) inclined surchnqe? 14.15 Describe the effect of water in the backfill on the active earth pressure on a retaining wall. 14.16 What is the pumpose of a shear key? Describe its action. 14.17 Briefly dcscribe the bchaviour of the various elements of a cunrilever retaining : wall.

\

746 REINFORCED

CONCRETE

DESlGN OF FOOTINGS AND RETAlNlNG

DESIGN

14.18 Briefly describe the beliaviour of the various elements of a countejfon retaining wall. 14.19 Where are the critical sections for shear located in the case of (a) the to; slab (b) the heel slab in the design of t l ~ ebase slab of a cantilever retaining will?: PROBLEMS 14.1 'Design a plain concrete footing for a column, 400 mni x 400 mm,curying an axial (service) load of 400 kii. Assume an allowable soil pressu~e.~oi 350 kN/mZ at a depth of 1.0 m below ground. Assume M 20 concrete and Fe 415 steel. 14.2 Design a square footing for a rectangular coluin~i 300mm x 500mm, reinforced with 6-25 $bars, and carrying a scrvice load of 1250 kN. Assume soil with an allowable pressure of 200 k ~ / n ?at a depth of 1.25 m below ground. Assume Fe415 grade steel lor both column and footing, and M 20 gmdc concrete for the footing and M 25 grade concrete for the column. 14.3 Repeat Problem 14.2, considering a uniaxial momcnt (with respect to the major axis of the column) of LOO !dim (under servicc loads - dead plus Live) in addition to the axial force of 1250 kN at the c o l u ~ nbase. Assume a snitable rectangular footing. Also assume that the moment is irreversible. 14.4 Design a square footing for a circula column, 500 nun in diameter, reinforced with 8-25 4 bars, and cai~yingan axial load of 2500 kN. Assume soil with a safe bearing capacity of 300 k ~ / m nat~a depth of 1.5 m below ground. Assnme Fe 415 grade steel for both column and footing, and M 20 grade concrete for the footing and M 30 grade concrete for Lhe colunm. 14.5 Repeat Problem 14.4, considering a rectangular footing with a spatial restrictionof 2.5 m on one of the plan dimensions. 14.6 Design a footing for a 250 mm thick reinforced concrete wall which supports a load (inclusive of self-weight) of 250 kN/m under service loads. Assurne a safe soil bearing capacity of 180 k ~ / mat ~a depth of 1 rn below gmund. Assunie M 20 grade concrete and Fe 415 grade steel for both wall and footing. Assunie the longitndinal reinforcement of the wall to comprise 0.25 percent of the gross cross-sectional m a . 14.7 Rcpeat Problcni 14.6, considering the wall to be made of niasonry (instead of reinforced concrcte). 14.8 Repeat Prohlem14.6. considering a bending momcnt of 30 kNm/m (reversible) at the base of the wall, in addition to the axial load of 250 kN/rn, under service loads. 14.9 Repeat the design of the two-colnnui combined footing of Exan~ple14.7, considering the property line to be located 500 mm away from the centre of column C 1

WALLS 747

14.10 Repeat the design of the two-iolulnn combined footing of Example 14.7, considering a beam-slab footing, and assuining that the allowable soil pressure is 180 kNlni2 (instead of 240 !&/in2). 14.11 Design and detail the sLem and basc slab of the cantilever retaining wall of Example 14.8. 14.12' De'sign a cantilever wall to retain earth with a backfill sloped at 20' to the horizontal. The top of the wall is 5.5 in abovc the ground level, and the foulidation depth may be takcn as 1.2 m bclow ground level, with a safe bearing capacity of 120 k~11n'. Assume that the bacUll has a unit weight of 17 kN/m2 and an angle of shearing resistance of 35'. Further, assume a coefficient of friction between soil and concrete, fi = 0.55. Use M 20 concrete and Fe 415 steel. . 14.13 Repeat Problem 14.12, co~rsideringthe backfill to be level, with a surcharge, equivalent to an additional 2.52 m of the backfill. 14.14 Suggest suitable proponions for a counterfort retaining wall to support difference in ground elevation of 9 m. Thc foundation depth may be taken as 1.5 m.below ground levcl, with a safe b e a h g capacity of 160 k~lm'. Assume a level bacldill with a unit weight of 16 kN/m3 and an anglc of shearing resistance of 30". Assume a coefficient of lriction, p = 0.5, between soil and concrete. Check the stability of the wall. 14.15 Design and dctail the various elements of the counterfort wall structure of Problem 14.14. REFERENCES 14.1

- E,xpla~mro,-yHn,trN,ook or, htdian Sla,rda~'dCode of Pruclice for Plairr and Reinforced Concrete (IS 456:1978), Special Publication SP:24, Bureau of

Indian Standanls, New Dclhi, 1983. Bowlcs, J.E., Fouridatiorls A,mlysis and Design, Thud edition., Mc~raw- ill Book Co., New York, 1982. 14.3 Peck, R.R., Hanson, W.E.. and Thombunl, T.H., Foundatiorr Engirzeering, Second edition, John Wiley & Sons Inc., 1974. 14.4 - Code of Proclicc for Sa-ttcruml Safety of Bl~ildirlgs:Shallow Fo~~nrlntions, IS;1904 (thin1 revidon), Bureau of Indian Standards, New Delhi, 1986. 14.5 - Code of Prrrcricc for Design and Co,rstrucfion of Simple Spread Fourdatior~s,IS:1080 (First revision), Bureau of Indian Standards. New Delhi, 1980. 14.6 ACI Comrriltce 336, Strggesrcrl Design Procerlrwrr for Conrbir?ed Footings and Mats, Journal ACI, Val. 63, No. 10, Oct. 1966, pp 1041-1057. 14.7 Kmmisch, F. and Robcrls, P., Sirrtpli/ird Design of Co,,hi,zeil Foori,,gs. ASCE Joulnal, Soil Mech;uiics Div., Vol. 87, NoSMS, October 1961, pp 19-44. 14.8 Huntington, W.C., Ear.rh Pressures nnd Retaining Walls, John Wiley, New York. 1968. 14.2

I'

748

REINFORCED CONCRETE DESIGN

Fisher, G.P. and Mains, R.M., Sliding Stability of Reruinbrg Walls, Civil Engineering, July 1952, pp 490. 14.10 Wang, C-K. and Salmon, C.G., X e i n f o ~ e dConcrete Design, Fourth edition, Harper & Row, New York, 1985. 14.1 1 - Code ofP,nctice for Conclere Structures for tl~eSlomge of Liquids, Part 4: Design Tables, IS:3370 (Part 4) 904 (Third revision), Burcau of Indian Staodanls. New Delhi, 1967.

14.9

15.1 INTRODUCTION The objective of structural design and constructiou is to build safe, serviceable, econortticol, d ~ ~ , o b l and e uesrheric structures. A~~alysis and design, together, comprise only one of the phases in the process of a building construction. Thc elaborate computations involved in this phase becorne worthwhile only if the design is translated into a correspondingly high quality svucture. This necessitales good detailing and construction practices. In Chapter 3, it was explained that the primary aim of design by the Limit States Method (LSM) is to minimise the probability of failure to an acceptable low value. In this context, fail~irv is dcfined as the attainment of a limit smre. Limit srates imply those conditions whereby a structure ceases to fulfil the functions [or which it has bccn designed. The limit statcs include both rrltirnate limit states and se,viceabiliry limit states. Thus, thc tcrm, failure, in general, includes both rdtirnntefiilu,u - local or overall - (exceeding the load carrying capacity, instability and buckling, overturning, sliding, fatigue and fracture, and pr.ogressive type of collapse) under factored loads, and serviceability failure (unacceptable deflections, vibrations, cracking, inadequate durability, permanent deformation, leakage, wetting, spalliug of concrete, etc.) under service loads. It is rarely that buildings fail in a manner that can be classified as an ultirnate limit state failurc (collapse). On the other hand, it is nluch too common for comfort that buildings (especially those of more recent construction) perform ~~~isatisfactorily in their day-to-day nonnal service; i.e., fail to meet serviceability criteria. In the fifties and sixties, reinforced concrete buildings used to be designed by the working stress method (WSM) with relatively low permissible stresses (for example, 5 MPa for 1:2:4 nominal mix concrete and 140 MPa for mild steel). Moreover, many analysis and design methods used in those days employed approximations which

GOOD DETAILING AND

750 REINFORCED CONCRETE DESIGN

'errcd on the safe side'. Modern structures are designed with higher strength matcrials, for higher stresses, and by the Limit States Method (including allowances for inelasticity and moment redistribution) wilh 'partial safety factors' lower than the 'factors of safety' inherent in the WSM as practised in the early sixties. l'urthermore, the methods of analysis and design have become more sophisticated and accurate, and thc conservatism in-built in approximate methods is no longer available., As a result, these modern structures am comparatively taller and have longcr spans, more slender members and thinner slabs and walls, and are built at a faster pace. They are therefore, more flexible (in terms of deflections) and are more 'crack prone', as compared with the old structures, which used to be low in height, had thicker (stockier) members, were lightly stressed and were built at a slow pace. Thus, the se~viceabilitycriteria assume far greater importance in modern structures. It is in this context that this chapter is included in this book. It is meant to draw the attention of enginccrs involved in all the stages of planning, design, detailing, fabrication and construction to thesc important aspects related to the perfortnancc of buildings. Detailing practices, construction practices, quality control in construction, building failures, causes arid prevention of crackdleakage in buildings, etc. are all large cnough topics to write separate. books and/or publish journ& on each of them, as indced have been done (Rei 15.1 to 15.8). Nor are all these topics strictly within the scope of this book. As such, an attempt is made here only to draw attention to some of the major and most colmnon causes of failure pertaining to the design (and construction) of reinforced concrete buildings. These are by no means exhaustive. For a more comprehensive coverage, reference may be made to Ref. 15.1-15.1 I. In this context it is worthwhile to note that, in most cases, the cost of the structure jtself forms only a small part of the total cost of a project. Further, the cost of the boncrete and reinforcernetlt rornls only a fraction of the cost of the stmctore. I-lcnce, the designer would do well to remember that:

15.1.1 Serviceability Failures

l'hc co~mnonlyobserved shortcomings in the context of serviceability requirements 81%: leakage from mofs and floors, particularly at construction / expansion joints, junctions, etc.; wetting of ceilings, walls (especially around toilet areas), leading to dampness, discoloration, growth of algac and moss on surfaces, growth of vegetation in fissures in walls and around drain pipes, sunshades, ledges, etc. 0 cracking in slabs, beams, walls, etc.; poor draimge and ponding of water on roof slabs, sunshades, bathrooms, open staircase steps, etc.; corrosion of reinforcement and spalling of concrete; e excessive deflections of slabs, beams, etc.

CONSTRUCTION

PRACTICES

751

Many of the above effects are interactive. For example, large deflections of roof slabs can lead to ponding of rain water as wcll as cracking on the top side in ~~egatlve moment regions, which in turn can lead to wetting and leakage, and also corrosion of reinforcement. 15.1.2 Reasons for Building Failures

I11

There are many causes that could lead to the failurc (ultbnnre and/or servicenbility) of a strmture. Some of tliese, which must be of concern to the dcsign and construction engineers are listed below:

I1

e

. . .

Deficiency in designlshift from actual design;

Poor detailing;

.

Failurc of a primary load carrying membcr by accident; Change in use (changc of structural arrangement) or overloading.

Poor quality mateiials;

Unforeseen disasters like sevcre earthquake, bomb blast, etc.

Poor quality construction.

Deterioration arising out of poor quality materials, construction and/or lack of repair and maintenance.

Poor formworWscaffoldi~ig

.

Exposure to adverse environment, not considered in original desigo

15.1.3 Structural Integrity Some causes for failures long aner completion of comtruction are identified in the above scction: Most of these causes such as accidents, overloading and disasters are not directly related to either thc design or the construction However, a related design consideration is the need for the structure to have stracta,nl integrity. A structure is said to have structural integrity if it is able to withstand localised damage or failure of a structural member, caused by any unforeseen or abnonnal events (that may reasonably be expectern without spread of damage or collapse to a large part of the stnature. In other words, the failure of one element should not lead to aprogressive collapse or inormental collnpse of the rest of the structure. A typical example of a progressive type of collapse is the failurc of a flat plate structure originating from the punching shear failure at one slab-coluilu~conneclion (in the absencc of special reinforcement for structural integrity at such conncctions; refer Section 11.7, Fig. 11.40). Punclting shear failure at w e columlt can lead to increased shear and momcnl at an adjacent column collnection, causing it to fail. Such progressive failurc of slab-column joints may lead to the slab falling on to the slab below, causing it to fail as well, and so on vertically down the building. Exenples of such progressive collapse are reviewed in Refs. 15.12 and 15.13.

!

752 REINFORCE0 CONCRETE DESIGN

In design, consideration should be given to the integrity of the ovcrall structural system to minimise the likelihood of progressive collapse. This involves a c a r e f ~ l selection of thc structural system, understanding its behaviour ondcr load and possible failurc modcs anrl cnsuring n robust and stable design . with sufficient rcrlrr,zrlnr~cyand dr@rnnrivcl o n d p n r l ~Most continuously reinlorced cast-inplace coocrete structures desirncrl - anrl detailed in acconlance with codes will "eenerallv, .iuossess a satisfactorv levcl of structural integrity. Special provisions for strocti~ral integrity may be lequircd for two-way llat slabs at slab-column connections, prccast concrcte suoctores, unusr!al slrnctural systems, and structures exposed to severe loads such as vehicle impacl, fire accident or explosion 15.2

DESIGN AND DETAILING PRACTICES

It is very rare that a building fails as a result of a major design flaw. The design codes are fairly consewative, if not up to date, as far as leinfolrcd concrete design is concerned. The engineering curric~~lum is also reasonably up to datc in this regard. Moreover, highly sophisticated and accurate softwares are now available for computer aided analysis, and also for design and draftinr. - However. in usine" these softwares, a word of caution is appropriate. The o u l p ~ ~01 t s rhesc programs are only as good as the inputs are! Hence they should only be used by persons who have a 1~111 understanding of what the program does, as well as 01 the propcrties of matedals, structural behaviour, failure modes, slructt~ralsystem employed, overall deformation patterns, compatibility conditions, load transmission paths to supports, and possible weak links, if any. Furthermore, it should be possible to identify the critical and primary load carrying elements, and to perform an approximate manual check on the stress resultants and design . resistance of such members, in order to avoid eross ewors. For example, the usc of appropriatc moment coefficients applied to a simplified substitute frame will yield a rough estimate of moments [refer Chapter 91. Similarly, an approximate estimation of the ultimate moment of resistance of a beamlslab section can be obtained as (0.87 fYA,,)(0.9d). As mentioned earlier, thc analysis of structures for stress resultants and the design of individual elements (critical sections of slabs, beams and colunins) for maximum load effects (bending moment, shear and torsion, and axial force) are done, in general, fairly competently. However, the attention given to the combining of these elements together to form the whole structure is generally found wanting both in the engineering curriculum and in many design offices. This includes the attention given to such important details as: termination, extending- and bending . of bars; anchorage . and development: stirrup anchorage: splices; construction details at connections (slabbeam, beam-column. rieid frame corncrs.. etc.): ...orovision of continuitv/discontinuitv at junctions of mcmbcrs; construction sequencing and minforccment placement to suit; deflection calculations (including long-term deflections) and control: crack control; special cases such as upturned (inverted) beams, cdge and spandrel beams, cantilevered nrernbers; cover, bilr support a d reinforcement protection; durability; recognition of and allowance for long-term effects of creep, shrinkage and temperature: details of contrbllconstruction/expansion joints; structures needing special procedures (tanks, chimneys, etc.).

GOOD DETAILING AND

CONSTRUCTION PRACTICES 753

The practical work in concrete Laboratory Courses in most universities is also found wanting. Experiments and assignments are mostly limited to the standard tests on cement, aggregatc and hardened concrete. Applied problems aimed at the understanding of concrcte technology (influence of various parameters, and concretc mix design) and fabrication and tcsting of rcinforced concrete elements (designfabricate-cast-cure-test-analyse type assignments) are seldom included in the laboratory courses. 1 5 . 2 3 Reinforcement Layout Heavy live loads and lack of regularity in framing (widely differing adjoining spans, variation in column sizes and spacings, flochlations in relative stiffnesses, etc.) can move the zone of contraflexure considerably, thereby affecting the termination, extension and bending of bars considerably. Such lack of regularity could necessitate continuing of top bars over the full length of a short span located between two long ones (common in school buildings, - hotels, etc. with narrow central corridors), o r may make it impracticable to bend any bar at all, or conversely, may make it possible to bend on considerablv more than half the bottom bars in a heavy . -pirder, ancborine.- and terminating them in a compression zone, or may necessitate provision of stirrups for the full length of a member. Admittedly, considerably improved design skill is retquired in delineating the reinforcement throughout the length of members than in selecting top and bottom bars for maximum moments. Concrete being weak in tension, reinforcement is provided to take care of all tensions envisioned by the designer, whether directlflexural (main reinforcement) or diagonal (stirrups). In addition, suitable reinforcement must be provided across any potential crack. In particular, attention should be paid to locations at which tensile stresses, not ordinarily calculated, exist - such as due to shrinkage, settlement, temperature and stress concentration effects. Cracking due to thermal and shrinkage movements can be reduced by making provision in the design and construction of structures for unrestrained movement of parts, wherever feasible, by introducing movement joints (expansion joints, control joints and slip joints). Where provision of movement ioints is not structurally feasible (as in rigid frames, shell roofs), thermal stresses have to be taken into account at the structural design itself. Even where joints for movement are orovided. some amount of restraint to movement due to bond and friction is unavoidable. In cases where the design reinforcement is only in one direction (as in one-way slabs, cantilevered slabs, etc.) cracks could develop across the perpendicular direction due to contraction and shrinkage in that direction, and it is necessary to provide some reinforcement (variously called as 'distributor reinforcement' or 'temperature reinforcement') in the direction perpendicular to the main reinforcement In case of members exposed to the sun (sunshades, fins. canopies, balconies, roof slab without adequate insulation cover, etc.), the "minimum" reinforcement soecified bv the Code should be increased by 50 to 100 pcrcent, depending upon the severity of exposure, size of member and local conditions [Ref. 15.51. Reference may also be made to Section 5.2.2, with regard to the need for side face reinforcement to control cracking in large unreinforced exposed faces of concrete members. ~

~

754 REINFORCED CONCRETE DESIGN

15.2.2 Design Drawings

Much time and expense can be saved, and costly mistakes avoided, if siinplc, clear and complete drawings are prepared. More than serving as a graphic dclineation of a structure, a drawing acts as a d e h i t e order to workmen to perform certain operations in a specified manner. The drawing also serves as a record of some of the important assu~nptionsmade in the design (which, for example, can reveal whether or not future expansion of the structure is possible a1 a later date). Engineering drawings prepared by the designer should specify grades of concrete and steel, live load, dimensions, reinforcement, lap lengths, concrete cover, and all other information needed for detailing the reinforcement, building the fornls, placing the reinforcement and placing the concrete. The designer has full data on the assuniptions made, the con~putations, moment diagrams and the whole philosophy of structural design, and it is his respo~?sibilityto define the design requirements by way of anchorages, laps, bends, splices and similar details. Indced, it is only he who can supply such information to the detailers, fabricators and construction personnel. Hence, design drawings must be complete to the extent that every bit of information regarding thc size and arrangement of concrete mcmbers, and the size, positioning and dctailing of reinforcing bars is completely covered either by a drawing, description, diagram, 'note, rule, or referencc to a standard manual. Notes and statements should be clear and unambiguous. A note "16 bars 2 ways 2 faces" is highly ambiguous, as it could mean 16 hars each way each face (total 64) or 4 bars each way 9ach face (total 16), or indeed almost anything in between. Instead, 'the note should hive been made explicit, giving the number each way in each face. imilarly, descriptions are best given in the imperative: "Do this", "Bend these bars", ther than "This may be done", or "These bars may be bent". Graphical representations (true-scale elevations, scctions, etc.) are preferable to coinplicatcd notes and descriptions, to show precisely what is wanted; they are also, to a considerable extent, self-checking. While, with uniform loads and equal spans, it may be satisfactory to bend bars at the quarter-point and to extend them to the 3110'~point as in Fig. 5.5 (following standard practiceslmanuals for such cases), this is not safe as'a general practice for all cases. Hence, the drawing should make it clear, by way of a separate section, indicating the departure from standard practice. '

15.2.3 Constructlon Details at Connections and Special Situations Frequently, locations and members needing special detailing considerations a1.e encountered. These include locations of abrupt changes in section size and other sudden discontinuities, edge beams, inverted beams, odd-shapedlsized members, connections, etc. A few such cases are described below. In many such instances, a uscful procedure for design andlor detailing is to adopt the strut-and-tie model [see Section 17.21.

GOOD DETAILING AND

CONSTRUCTION PRACTICES 755

Offset Columns When a column in a particular storey is smaller than the one below, some of thc vertical bars from below may have to be offset to come withi11the column above, or dowel splices must be uscrl [Fig. 15.l(a)l. 'The slopc of the inclincd portion should not cxceed 1 in 6. Wherc column verticals are offset, additional ties shall be provided and placed close to tlle point of bend in order to carry the transversc force generated due to the chaoge of dircction at the bend. When the offset between column faces exceeds 75 nun, the vcrtical bars in the column below shall be terminated at the floor slab, and splicing of colullln bars by dowels may be necessary [Fig. 15.l(b)]. Dowels may also be necessary when the placing of part of the structure is delayed, and also between various units of structures (such as footings and columns). Dowels should, in general, be of the same size and grade as tlle hars joined, and should be of sufficient length to splice with the main oars. When column bars arc spliced, additional ties shall be provided at and near the ends of spliced bars, to pmvidc confinement to the highly strcssed concrete In the regions of the bar ends.

Members with a Break in Direction Whenever there is a changc of direction in a main reinforcing bar, a resultant radial force is generated at the location of the kink, as shown in Fig. 15.2(a). If the radial force acts outwards, as is the case in Fig. 15.2(a), this force tends to pus11 out the cover concrete causing splitting. Moreover, as a straight length (such as AC) is shorter than the bent length (ABC) of the bar, the spalling will lead to a relieving of the bar stress resulting in lowering of the resistance of the section, and possihle failure. When the angular changc is small (say < 15') the radial force msultant ( K ) is small and can he cai~iedand transferred to the compressio~lzone by providing ademate number of stirrulx at the location of the kink and on either side at close --.7..... spacing as shown [Fig. 15.2(b)l. When the angular change is larger, the reinforce~nentfrom eithcr side should be continued straight and anchored lo develop the full design stress isec Fig. 15.31. Note that in Fig. 15.3 the bar A, which cannot get a straight length of Ld beyond the location of the kink in the beam face is continued on the compression face and anchored there, so that no outward splitting force is developed due to the bend in this bar. Examples are junction of stairs and landing' , inside comers of rigid frames [Fig. 15.4(a)j and wherc the soffit of beam forms an angle as in n gable bent [Fig. 15.4(b)l ~

~~~~

'Tlis detail is indicalcd in Fig. I24(n) of Chapter 12.

756 REINFORCED CONCRETE

GOOD DETAILING AND

DESIGN

CONSTRUCTION PRACTICES 757

(SECTION A- A

lower bar

(b)

(a)

Fig. 15.2 Member with a change in d~rectionin flexural stresses (small angle)

Fig. 15.3 Large directional change in flexural stresses

inside bars to be xtended separately bottom bars to be

(a)

lap as specified

(W

" 4

Fig. 15.4 Reentrant corners with tension bars COnStr~Ctio joint

Fig. 15.1 Some construction details at connections

A similar situation exists when the internd co~npmssionforce changes direction i n such a way that the resultant force acts outward [Fig. lS.S(a)]. I n the example shown, a breaking away of the flange can be prevented by transverse i-einforcement tying the flange to the web of the beam [Rg. lS.S(b)]. Construction and bar placing details of the corner connection o f a rigid frame are shown in Fig 15.l(c). In detailing such connections, care must bc taken particularly i n providing full continuity around as large a uniform radius as possible i n splicing

GOOD DETAILING AND

758 REINFORCED CONCRETE DESIGN

the top bars from the girder to the outside bars in the colunul. Rigid comer connections of beams to columns often requjre closed stirrups or tics around the bend [see also Section 15.2.4 on Rigid Frame Joints]. The designer must provide complete information showing the radius of bend, location and dimensions of the lap splices (or other type of splices) used and sdrrup details.

CONSTRUCTION PRACTICES 759

Edge Beams In edge and spandrel beams, stirrups must be of the closed type and at least one longitudinal bar should be located at each corner of the bcam section. Typical details are shown, for 11ornla1 and inverted edgelspandrel beams in Fig. 15.6(a) and (b) respectively. For easicr placing of the longitudinal bars in an inverted beam, twopiece closed stirrups can also bc used as shown,in Fig. 15.6(b).

Corners of Walls In concrete walls, horizontal reinforcement may be required to resist moment, shear o r temperature and shrinkage effects. All such bars in both faces of wall must be sufficiently extended pas1 a corner or intersection to satisfy development requirements. Typical details are shown in Fig. 15.7 for resistance against molnent (inward and outward), with the reinforcement from the approp~iatefaces anchorcd. (a)

(b)

not less than 12 0.300 mm or 50% of lap length specified

not less lhan 24 B, 300 mm 0 , 50% of lap length specified

Fig. 15.5 Change in direction of compressive stresses

optional to goohook,

/

closed by standard 9O0stirrup hook ax+an+inn=- 6R 0n hook, extension

.. all stirrups provided in edg beams must be oiosed

(a) stirrup forming closed tte

minimum 10 6 bars. Ontinuous, except when spliced to other top stesi. m u d ha These bars must be same same size as stirrups if stirrups are largerthan 10 0

corner bars must be ,oronarlv --- , anchoredat supports

Fig. 15.7 Typical corner details in Walls

Speclal Conditions minimum 10 C$ bar Continuous except w Spliced to Other tops CQnstructionjoint where

forming closed tie Straight bar splice; lap length as specified by designer colner bars must be properly anchored at suppons

(b) two-piece stirrup forming closed tie

Fig. 15.6 Typical edge i n d spandrel beam details

For special or unusual conditions, adequate details should be show11 for proper placing of reinforcement, as the average steel setter cannot be expected to understand engineering principles. Examples are cantilevers and continuous footings, in which the reinforcement is in the opposite side from the one to which the steel setter is accustomed. There are situations a l w e the embedment length available for end anchorage of bars is insufficient to dcvelop the design stress in the bar through bond. Examples include corbels, deep beams, snl?ll size footings, precast beams, ctc. In such cases, special devices such as welded cross bars, end plates [Fig. 15.81 must bc provided.

GOOD DETAILING AND

760 REINFORCED CONCRETE DESIGN

CONSTRUCTION PRACTICES 761

angle end plate (welded)

end plate /(welded to bars primary beam

3

&nary bea; (girder)

Fig. 15.8 Special anchorage devices Intersection of Members Congestion of steel should be avoided at locations where members intersect, such as intersection of (secondary) beams with girdcr (primary beam) and girders with colu~mi. In the interior beam-colunu~joint, generally thcrc is o v ~ r c r o w d i n ~ othc f negative (top) reinforcement in the beam if they are all placed within the beam widlli [Fig. 15.9(a)l. This usually interferes with proper placing and compaction of the concrete at the joint. Thc bond developed in these top rei~forcementalso is relatively inferior. The spreading of thc top reinforcement into the adjoining slab, preferably using smaller diameter bars, [Fig. 15.9(b)l has been shown to reduce thc crack-widths in these beams considerably [Ref. 15.121. This has the added benefit that the effective depth is slightly increased and the placement and compaction of concrete is facilitated better.

Fig. 15.10 Bars of secondary beam lo be place0 over oafs of pillwar). boam, wth sdspender st rrdps enclosing prlmaty boatn oars 15.2.4 Beam and Column Joints (Rigid Frame Joints)

Joints of beams and columns in rigid-jointed frames are critical locations requiring careful design and detailing. Such joints should have adequate strcngtli to enablc tile developmcnt, at the joint face, of the full design strengths (and plastic hinges with adequate ductility, if required by design) in the members framing into eachjoint ltnder the most adverse loading pattern, without distress in the joint itself. This type of rigid coimection occurs in rigid jointed multistorey frames, portal frames, box culverts, at the base of cantilever retaining walls, etc.

- M

(a)

Fig. 15.9 Negative moment reinforcement at beam-column joint At the intersection of a beam and girder, the beam bars should be placed at a different elevation than those in the girder so as to avoid interference. The relative positions of the bars must be in accordance with the load transfer order assumed in design. Thus the beam reinforcement must come over the girder rcinforcement at the intersection [Fig. 15.101. In addition, adequate hanging up bars (suspender stirrups) should be providcd in both members in the joint zone as given in Section 6.10 [see also Fig. 6.71. Similarly, at slab-beam junctions, the bottom and top bars of the slab must be draped over the bottom and top bars in beam respectively. When slabs frame flush with the bottom of inverted beams or hanger walls, special stitrup hanger reinforcement shall be provided [see also Section 6.5 and Figs 6.7 and 15.41,

(b) dlagonal compression and

(c) tensile strains along

possible splitting cracks

diagonal plane closed ties in both direclions

steel bars for crack control

(d) joint reinforcement for

large size joints

(e) alternative arrangement (especially under moment reversal)

Flg. 15.11 Knee joint subjected to 'closing' moment

A simple example of a rigid beam-column joint, for which the design and detailing considerations can be explained with relative ease, is the corner joint of a portal frame, usually called a "knee joint" [Fig. 15.1 I]. The joint, to be rigid, must have full continuity between the two mnembcrs. Thc main flexural reinhrcement in the joint zone undergoes a change of direction, and as a result transverse forces are dcveloped, as in the cases considered in Fig. 15.11. There can be three types of stress pattcrns to which the joint zone is subjected to, depending on the nature of loading on the structure itself, namely: (a) where the moment at the joint tends to 'close' the knee i.e., hogging moment causing tension in the outer fibres [Fig. 15.11], (b) where the moment tends to "open" the knee [Fig. 15.121 and (c) where the lnoment is subject to reversal as in the case of seismic loading. The nature of induced forces in the joint zone in case (a) is shown in Fig. 15.11(b) and (c). Thc diagonal resultant thrust tends to develop splitting cracks along the diagonal ac. A significant portion (ae) of the diagonal ac will be under tensile stress and liable to crack, as shown in Fig. 15.11(c). It may be noted that thc joint is usually subjected to axial forces and shearing forces, in addition to thc bending moment. For satisfactory performance in this case, the outcr tension bars should be continuous around the corner. The inner bars are in compression; however, the concrete alone may be adequate to cairy the compressive forces here. These bars are better continued straight, as shown, rather than being made continuous by bending around near there-entrant corner. In case these bars are also accounted as contributing p a t of the required compressive force, they should be continued straight and anchored to dzvelop the design stress at the joint faces along corner c. Furthermore, the diagonal compression along ac and the possible diagonal cracking along ae should be countercd. When the members are of small size (as in the case of slab-wall joint or the corner of a small size lightly reinforced and thin walled box culvert), no special provisions may be needed to carry the diagonal compression and tension along diagonals ac and bd. However, in large size or heavier reinforced members, the diagonal compression may be resisted, and the diagonal cracking controlled, by secondary reinforcements placed along diagonal directions as shown in Fig. 15.1 1(d). An alternative arrangement is to place these reinforceme~lt orthogonally as in Fig. 15.1 1(e). This arrangement is particularly suited for cases of moment reversals. The case of a knee joint subjected to 'opening' moment is shown in Fig. 15.12. In general this loading case (moment tending to open the knee joint) is more critical than the case of moment tending to 'close' the knee joint [case (a) discussed in Fig.15.1 I]. The nature of stresses in the joint and potential crack locations are shown in Fig. 15.12(b) and (c). The need to provide reinforcements along diagonal ac to carry the resultant tension in this direction and parallel to diagonal bd closer to the interior corner c, to control flexural cracking are self evident. The concrete near the comer a is stress-free and is likely to spall off, being pushed out by the resultant thrust along ca, and nominal reinforcement is required to control such cracking. Suggested reinforcing details [Ref. 15.111 for large size joints of type (b) are shown in Fig. 15.12(d).

V

diagonal cracks

(a)

(b) resultant tension

along diagonal

(c) flexural stresses along diagonal

(d) details of reinforcement

Fig. 15.12 Knee joint subjected to 'opening' moment

When the moment is subject to reversals, the concmte in the joint zone is likely to crack along both diagonals and significant anlounts of seco~ldaryreinforcements We required along both diagonals. For this situation it is more convenient to providc an orthogonal mesh of reinforcement (horizontal and vertical) in the joint zone. in the form of closed ties. These will resist the horizontal and vertical components of the tensile forces along the diagonal. A model for computing the area of the horizontal and vertical secondary steel (stirrups) requircd is suggested in Rcf. 15.1 1. When the joint is subject to high intensity reversed loading for several cycles, the concrete in the joint is likely to be cracked along both principal directions, and it is recommended that the resistance offered by the concrete should nut be Vdken into account. In multi-storey building frames the joint behaviour is more co~nplexas up to four beams may be framing into a joint with columns above and below at an interior joint. When both beams framing into a joint from opposite directions rcach thcir ultimate capacity and bend in a revcrse curvature mode, the diagonal co~nprcssiveand tensile stresses induced in the joint panel may be very high. Moreover, thc beam!column reinforcement may h e to develop full anchorage within the joint zone, which may be difficult if the concrcte is scverely cracked, parallel to both diagonal directions. The joint shear may also be twice as high as that in an exteuio~joint with beam only on one side. The diagonal compressive stresses and potential diagonal cracking would require an orthogonal mesh of well anchored horizontal and vertical reinforcement ties in the joint region. Furthermore, the concrete in the joint core should be laterally confined. Effective lateral confinement would require cross ties between the legs of the orthogonal ties to prevent the lateral bulging of the concrete in the core. 15.2.5 Construction Joints It is desirable to indicate at l c ~ ssome t of the more obvious construction joints so that all the trades are working along the same line. This also facilitates the designer to indicate shouldered joints [such as in rigid frame bents - see Fig. 15.l(c)l, reinforced to take moment, shear and thrust. The lapping and splicing of bars should

764 REINFORCED CONCRETE DESIGN

GOOD DETAILING AND

CONSTRUCTION

PRACTICES

765

be illustrated clearly, and not just schematically (indicating the amount of steel required at a few points). 15.2.6 Bar S u p p o r t s a n d Cover It is essential to have the reinforcing stecl accurately located in the Corms and firmly held in place belorc and during the placing of concretc by means of supports and spacers. Such supports should be adcqoate to prcvent displacement during the course of construction and to keep the bars at the propcr distance (cover) from the forms. In countries such as USA and Canada, standard bar supports in the lorm of individual or continuous metal bar chairs (plain, galvanised, or plastic protected) are commercially available and usually specified, although precast concretc blocks am also used. Howexr, in India, while the recomnended practice is to use precast mortar blocks (of thickness equal to the specilied covcr) lor bottom bars and individual (locally fabricated) mnctal chairs for tophent bars, actual site practices vary considerably. It is fairly conniion to see bottom bars being supported by just slipping in pieces of coarsc aggregate betwccn the bar and the formwork. Ncedlcss to say, tliesc angular aggregate pieces, precariously poised between the Cormwork and thc round stecl bars, slip out cnsily during placing of concrete (if not earlier). The rcsult is that thc bar often comes to rest on the lormwork with littlc or no covcr. When the forniwork is removed, it is a sorry sight to see the bars exposcd from underneath slabs and beams at several places! This gets covered up in plaster soon enough to givc an appearance that all is well (well, at least for the time being!). However, it is not long before the poorly protected reinforcing bars get conoded, and in this process increase in volume, setting up internal bursting stresses in the concrete. In course of time, this causes first cracks in line with the reinforcement, and later spalling of the concrete dislodging the plaster and whatever little concrete cover there is, thereby fully exposing the corroded bars. The seriousness of the resulting damage is obvious to all. In the case of top bars, locally made individual high chairs are used. Howcver, often these are few and far between, and at times are too flexible [fig. 15.131. In such cases, with the unskilled workmen frequently stepping on the top bars, the chairs may get bent (sag) or bent bars may get turned sideways, in either case resulting in a reduced effective depth (particularly so in slabs) in the negative moment regions. A reduced depth in this high moment region is likely to lead to undesirable cracking on the top face of slabs near the support (continuous) regions. This is one reason for the wetting visible underneath roof slabs where they join supporting beams. The multiple and cumulative effects of reduced effective depth at support on deflections, cracking, wetting, leakage, and corrosion of reinforcement can be casily ondcrstood. lnadeyrrnte concrete cover- for sreel bars is rr very comn~only observed co,wrucrio,i error in India. Much more attention needs to be paid for providing adequate cover and propcr bar supports. Significantly, the rccent (2000) revision of thc Code has enhanced the cover requirements for reinlorcements, including links.

Fig. 15.13 Inadequate rigidity of a steel chair

15.2.7 Deflection Control In Section 15.1, it was explained that modern designs result in relatively slender members with associated larger deflections. Creep and shrinkage causes deflections to incrcase with age, and such increases ntay be as much as 2 to 3 times the initial elastic rleflections. The possible adverse effects of lmge deflections o n a continuous roof slab is schenlatically shown in Fig. 15.14.

Flg. 15.14 Adverse effects of large deflections on a continuous roof slab Beam/slab deflections can also cause cracking in other elements (such as walls) sopported by it. This points to the need for deflection calculations and control. While for normal slabs and beams it is enough to control deflections indirectly by limiting spanldepth ratios [see Section 5.31, for members which are heavily loaded or exposed to adverse environmental conditions, it is necessary to calculate initial and long-term deflections and to ensure that these are within limits. In large spans where significant deflections can be anticipated, it is desirable to provide initial upward camber in floor slabheam so as to offset deflection, especially in roof slabs and cantilevered spans. 15.3 MATERIALS AND CONSTRUCTION PRACTICES Strict adherence to codes and specifications, use of good quality materials, engagement of trained and skilled masons and labour, and good workmanship under strict, honest and competent supervision are prerequisites for avoiding malfunction and failures, and for ensuring high quality structures. Regular materials testing, selection of competent contractors, supervisors and construction engineers as well as periodic inspection by regulatory agencies are needed for good quality construction.

766 REINFORCED CONCRETE

DESIGN

Poor quality of construction n~aterialsis a real problem in India. Periodic testing of building materials is seldom resorted to in most construction sites. Adulteration of cement has been rcported from the sites of even major hydm-electric projects undertaken by large public sector organisations. Reinforcing steel is being supplied by many rerolling mills in the small scale sector, which do not have in-house testing and quality control facilities, with thc rcsult that there is little guarantee about the strength, ductility, uniformity, and dimensional tolerances of such bars. All these underscore the need for periodic testing and quality control of materials used for constiuction. The major reason for poor performance of reinforced concrete structures is the poor quality of construction. This is apparent if one realises that with the same quantity of cement and aggregate, both very poor quality and very high quality concrete can be obtained, depending on the water-cement ratio used and degree of control exercised. With reduction in wlc ratio, nearly all the engineering properties of concrete (strength, modulus of elasticity, durability. reduced shrinkage and creep, impenncability, etc.) improve. Reduction in creep and shrinkage also results in reduced long-term curvatures and deflections due to them, and reduced shrinkage cracks in concrete. Yet, most masons routinely use cxccss water in the mix to save time and labour on compaction and screeding; and this could be a major contributii~g factor to evcry one of the serviceability failums listed in Section 15.1. It being the most important single factor influencing concrete quality, the quantity of'water used in the mix should be the minimum, consistent with requirements of laying and proper compaction. To enable the use of a dry mix (low wlc ratio) and yet get good compaction, all srrucru,al concrete should be compacted using vibrators There are other requirements for good quality concrete like grading and cleanliness of aggregates, weight-hatching, thorough mixing, adequate curing, etc., and for more details on these and other such requirements, reference may be made to books on concrete technology [Ref. 2.1-2.61. In this context, it is worth noting that leakage through concrete slabs is a very frequent problem in many parts of the country. Such leakages, if restricted to small areas, can be repaired. However, repairs of a poorly built roof slab with extensive leakage spread over large areas is virtually impossible'. Therefore, it is prudent and itpays to do the initial construction meticulously. Yet another common mistake seen usually in sites of small projects relates to roq flimsy and inadequately supported formwork. Such forms deflect and vibrate as workmen move about over it. Deflection and vibration during periods of placing, setting and early curing of concrete can result in cracks developing in the concrete. Formwork supports are also frequently seen to be infirm, unstable or yielding. Other common constructions errors include inadequate curing, bnproper levels and slopes inadequate drainage arrangentents, pool. bonding behveen harderred and

'

Note: Vanous manufactures and propritery agencies advenise various methods such as tar felting, special plastering with chemicalladhesive cements, etc., far repairing leaky slabs. These may work far small areas and for a shon period for larger areas, but do not offer a permanent solution.

GOOD DETAILING AND

CONSTRUCTION PRACTICES 767

frrshly laid concrete ar co,,stmcrion points, poorly construcrcd expa~tsionjoints, supe?ficialfilling of holes cut for.pl~artbbtg/electr~icnfio~~fixt~~re~s, etc. Reference has been made here to deterioration arising out of poor quality materials and construction. Associated with this is the need lor timely repair and maintenance. Apart from mgular and routine inspection and maintenance, every concrete structure should undergo a spccial inspection and associated special relmi,.; once in 10-15 years.

15.4 SUMMARY In this chapter, an attempt has bcen made to draw attentiorl to some of the essential precautions to be taken and to a few of the very common lnistakes with regwd to the design, detailing and construction of concrete structures. It would be desirable for designers to develop a list of "do's and don'ts" based on codes and specifications, manuals such as Ref. 10.1-10.6, and their own experiences and observations. A partial list of such guidelines is given bclow: It should be ensured (to tlic extent possible) that the materials spccificd can be rcadily obtained in the s i x , length and gradc required. The quality of materials should be ensured by regular materials testing. Apnrt from strength and durability considerations. the specification lor concrete should also be decidcd so as to obtain mhimmn of drying shrinkage and creep. Concrete mix design should aim at obtaining durable concrete of required strength through proper grading of aggregates, control of wlc ratio, thorough mixing, proper compaction and adequate curing. Thc quantity of watcr uscd in concrcte should be the ~ninimun~ practicable, consistent with requircments for proper placement and compaction. Consiste?t with requircments of economy, the integrity of thc structure should be imn~roved by building in structural redundancy in the framing system, reinfnrcemcnt placement, etc ~ l ~ members ~ ~ (slabs ~ a andl beams) should have adequate stifflll~sso as to limit deflections. In members liable to undergo large deflections, upward camber may be provided to offset the deflections. Reinforcement design and detailing should take care of all tensions in concrete, whether direct, flexural, diagonal or due to shrinkage and tempcratore. Suitable reinfnrcemcnt should be provided across all potential cracks in concretc, whatever the cause. To minimise shrinkage and temperalure stresses, wherevcr fcasible, provision mav be made for onrcstricted movements of parts, by introduci~~g control/expansio~l/slip joints. Concrete slabs in exnosed situations, such as sunshades, balconies, canopies, .~~ open verandas, etc. sl,ould hc provided with adequate quantities of temperature ~.einforcementin onlcr to prevent cracks due to shrinkage and contraction

768 REINFORCED

CONCRETE

GOOD DETAILING AND

DESIGN

Adequate develop~~icnt length andlor anchorage should be provided so that the conq~uterlstress at every section of a reinforcing bar is fully developed on both sides. It should be ensured that hooked and bent bars can be placed conve~iie~~tly and have adequate concrete protection Congestion of bars should be avoided at points where nicnlbcrs intcrsect, and it should be cnsured that all the reinforcement required can bc propc~lyplaced. When a member has a break in its direction so that tl~creinforcement in tetlsion tends to sepnratc from thc body of concrete, special anchorage should bc provided and properly detailed. When slabs frame flush with bottom of inverted beams or when a load is applied to the side of a member through brackets, ledges or cross beams, special stirrup hanger reinforcement should be provided. Liberal concrete cover for reinforcement should bc provided in general, and particularly in hunlid, wet or aggressive environments. The desired cover should be ensured in actual constmction with proper cover blocks 1 bar supports. Complete and accurate dimcusions should be specified in engineering drawings.

CONSTRUCTION PRACTICES 769

15.3 What are the factors in concrete-making that influence creep and shtinkage of the concrete ? 15.4 What are thc precautions to be taken (a) at the design stage, (b) at the detailing stage and (c) at the construction stage for ensuring ahigli quality structure ? 15.5 What are the types of serviceability faihrm that can occur ? 15.6 Describe a 'serviceability failure' that you have observed in a structurc you are familiar. with, and analyse its causes and effects and suggest remnediallrepair

measures. 15.7 Explain the concept of 'structural integrity'. 15.8 In a three-hour long training programme to be given to masons engaged in construction of concrete structures, list out the important topics that you would include. 15.9 Make a literature survey and wxite a 'state-of-the-art' report on (a) Building failure studies: (b) Deflection calculations and control m concrete structures; (c) Cracking and control in concrete structures; and (d) Leakage and control in concrete buildings.

REFERENCES clearly in drawings. Details of corners, intersection of members, control and coiistmction I expansion joints and similar special locations should be drawn. For special and unusual conditions, adequate details sliould be shown in drawings so as to ensure proper placing of reinforce~nent [Examples: cantilevers, continuous footings, hinged base of rigid framcs, etc.] Propcr bar supports should be provided to ensure that thc winforcing bars arc accurately and firmly held in place before and during concreting, and thus the required concrete cover and effective depths are obtained. Compaction of concrete shookl be done using vibrators (whcmvcr feasible), enabling the use of low wlc ratio and ensuring better strength and durability. Formwork should be built firmly and with rigid supports and without gaps and holes through which the cement oaste can escaoe. Curing of the concrete shoukl be done for durations as recormnended by the Code, and should be tcrminatetl gradually to prevent quick drying. In case of members which ace liable to large deflections (Examples: cantilever beams and slabs), the removal of centering should be delayed as much as possiblc so that the concrete attains suflicient strength. Adequate slopes for roofs, bathroom floors, etc., should be provided to ensure quick drainage.

15.1 15.2

15.3 15.4 15.5 15.6

-

REVIEW QUESTIONS 15.1 List the arcas in which the basic Reinforced Concrcle Design course content in your university is dericient. 15.2 List ten most common construction mistakes in the order of their importance.

Reinforcement and Derailing, Special Publication SP 34, Bureau of Indian Standards, New Delbi, 1987 -Concrete Materials andMethods of Concrete Corzstrucrion, CSA Standards A 23.1-94, Canadian Standards Association, Rexdale (Toronto), Canada, 1994. - Structuml Failures: Modes, Causes and Responsibilities, A~nerican Socicty of Civil Engineers, New York, 1973. Feld, Jacob, Construction Failures, John Wiley & Sons, New York, 1968. - Handbook on Causes and Prevention of Cracks in Rrcildings, Special Publications SP 25, Bureau of Indian Standards, New Delhi, 1984. - Fomwbrk Swiking Times - Criteria, Prediction and Methods of Assessment, CIRIA Report, American Society of Civil Engineers, New York, - Ifandbook on Concrete

I

nnl:

137".

Hoover, C.A. and Greene, M.R. (editors), Constr.irction Quality, Education and Seisrnic Safety, Earthquake Engineering Research Institute, Oakland, California, 1996. 15.8 Grant, E.L. and Leavenworth, R.S., Statisricul Qualip Corztral, Sixth edition, McGraw-Hill BookCo., New Yo~.k,1988. 15.9 Levin, R.I., Rubin, D.S., Stinson, J.P. and Garden Jr., E.S., Quantitutive Approaches to Management, Eighth edition, McGmw-Hill Book Co., 1992. 15.10 Wacstlund, G., Use of High-Slrength Steel in Reinforced Concrete, Journal ACI, Vol. 30, No. 12, June 1959, pp 1237-1250. 15.11 Park, R. and Paulay, T., Reinforced Concrete Strucrures, John Wiley & Sons, Inc., Ncw York, 1975. 15.7

.

770 REINFORCED CONCRETE DESIGN

15.12 Allen, D.E. and Shriever, W.R., Progressive CoNapse, ~ b ~b a d~s a,ld , ~ ~ l Building Codes. in 'Structural Failure: Modes, Causes, ~ ~ ~ ~ ~ ~ ~ ~ i b i l i t i ~ ~ s , Amellcan Society of Civil Engineers, 1973, pp 2 1 4 7 . 15.13 D.A.3 Progressive Collapse, Canadian Joulnal of Civil Engineering, Vol. 2, No. 4 .Dec. 1975, pp 517-529.

16.1 INTRODUCTION

During an earthquake, ground motions occur in a randon, fashion, both horizontally and vertically, in all directions mdiating from the epicentre. The ground accelerations cause structures to vibrate and induce inertial forces on them. Hence, structures in such locations need to bc suitably designed and detailed to ensurc stability, strength and serviceability with acceptable levels of safety under seisnric effects. The resultant inertial force at any floor levcl' depends on the mass at the floor level and also the height above the foundation. The inertial forces usually follow a parabolic valucs at the top floor distribution in rcgular nrulti-storey buildings, with n~axi~num levels. In regions of high seismic intensity, it is desirable to tninimise the weights at various floor levels, especially the roofs and upper storeys. Also, it is desirable to avoid discontinuities in mass or stiffness in plau or elevation. Torsional effects should particularly bc accounted for in buildings with asymmetry in plan'. The codes published by the Bureau of Indian Standards, which spcciry nrinin~um design requirements for ewthqunke-resistnnt design, are listed as Refs. 16.1-16.3. These requiremetits take into consideralon the characteristics and probability of occun'ence of earthquakes, the characteristics of the structue and the foundation, and the a~nount of damage that is considered tolerable. References 16.4-16.7 give details of code provisions in some seismic rcgions of the world.

' Sufficier:t number of n~odcsof vibration have to be considered is the 'response spectmm'

analysis, as prescribed in IS 1893 (2002). and n suitable luode combinarion scheme (such as 'SRSS' or 'CQC') has to be employed. Torsional effects should bu considered w h a ~the eccentricity between the 'centre of mass' and 'centre of stiffness' at any floor level is significant (more than 5% of the floor plan dinlension).

772

SPECIAL

REINFORCED CONCRETE DESIGN

The criteria adopted by codes for fixing the level of the dcsign scislnic loading are generally as follows [Ref. 16.71: structures should bc able to rcsist mirtor eartliquakes withoul damage: structures should be ablc to resist moderate earthquakes without significant structural damage, but with some nonstructural dnniagc; and e structures should be able to resist ,najor.carthquakcs without collapsc, but with sonle structural and nonstructural damage. The magnitude of the forccs induced in a structure duc lo a given ground acceleration (or given intensity of earthquake) will depend, amongst other things, on the mass of the structure, the material and type of co~~structio~l, and tbc darnping, rlrdlir), and enwgy rlirriporior, cnpucity of the structorc. By enhancing ductility and energy dissipation capacity in the structurc, the induced seismic forccs can be reduccd, and a more economical structorc obtained, or alternatively, thc probability of collepsc reduccrl. Buildings with latcral load resisting systcm co~nprising(i) a ducrile moazorr-r.r.sisrir~g.qx~cc,fr-rrmeor (ii) a dual systcm consisting of ductile moment rcsisting space frame and ductilc flexural (shear) wall, qoalil), for very low seismic induced forces.

relatively high intensity seismic zones (zones 111, IV and V), specified in IS 1893 : 2002 [Ref. 16.11.

.

,

.

D~uclil/tymay bc uroanly dcfinep as.1h.dao~lily,ofa s l r ~ c i w eoi,n~ember?to, ;uria,erqo inelasti~o~lor~nalions 4qyona,rtie,:n,l,at;y elcl (Iclor~~atloll w.lh,llO: .. :d&~r&e:in'.lhe load res,stanco. ..::.,. ; : .: .'; . _: .,. a,. :

PROVISIONS FOR EARTHQUAKE-RESISTANTDESIGN 773

16.2 IMPORTANCE O F DUCTILITY IN SEISMIC DESIGN 16.2.1 M e a s u r e s of Ductility A general qualitative definition of ductility was given in the preceding section. A qu.antitative measure of ductility has to bc with reference to a load-deformation ~-~~~ response. A d~rcrileresponse would be reflected in the deformation increasing a1 nearly constant load such as was shown in Fig. 9.8. Then, the ratio of the ultimate deformation to the deforrnatioli at the beginning of the horizontal path (or, at first 'yield') can give a measure of ductility. However, each choice of deforn~ation(strain, rvrariorr, curvrrtrrre, or dcpection) may give a different value for the ductility measure. ~~~~

Curvature Ductility For an under-reinforced beam section in flexure, the moment-curvature (M-q) relation is typically as shown in Fig. 16l(a). Based on the idealised M-v behaviour, cur-vortor ductility, p, may be defined as the ratio q,,/'p)., whcre vs is the curvature at first yield (idcalised), and v,,the maximum (ultimate) curvature ~t the section:

..

Since reinforced concrete is relatively less ductile in compression and shear, dissipation of seismic energy is best achieved by j?e.rrrr.nl )rielding. A frame of continuous constmction, comprising flexural members, CO~UINIS and their connections, designed and detailed to accommodate reversible lateral displacements after the formation of plastic hingcs (without decreasc in strength), is known as a; ductile moment-resisting f i m e . Similarly, shear walls (more appropriately called j7exu,nl walls), are reinforced concmte structural walls cantilevering vertically from. the foundation, and dcsigned and detailed to be ductile and to resist seismic forces and to dissipate energy through flexural yielding at one or more plastic hinges. Modem codes [Ref. 16.1t, 16.4, 16.51 provide for reduction of seismic forces through provision of special ductility requirements. Details lor achieving ductility in reinforced concrete structures are given in IS 13920 [Rel. 16.31. Methods of, determining design seismic forces, either in the form of equivalent static lateral loading or through proper dynamic analysis, lie outside the scope of this chapter; the reader may lefer to the codes, handbooks and other texts [Ref. 16.8-16.101 for this purpose. This chapter explains the major code provisions, particularly those given in Rcf 16.3, dcaling with designing and detailing for ductility in nlomcnt-resisting frames and shear walls. Such provisions are mandatory for structures located in

In the recent revision of IS 1893 (2002), the procedure recornm&ded is to first calculate the actual force that may be experienced by the structure during the 'probable maximum, earthquake', if it were to remain elastic. Then, the,effect of ductile dcfomntion and energy, dissipation is accounterl for by means of a 'response reduction factor'.

Indeed, IS 13920 [Ref. 16.31 defines curvnrure ductility as the ratio of curvature at the ultimate strength to the curvature at first yield of tension steel in the section. The value of p is a property of the beam crvss section, and can be computed easily using the principles described in Chapter 4. It can be shown that 'curvature ductility', U , of a section increases with:

A curvature ductility of at least 5 is considered to be adequate for reinforced concrete [Ref. 16.21. Different M e a s u r e s of Ductility In the case of a beam member [Fig. 16.l(b)l, it is more difficult to define a unique ductility ratio, as it could be in terms of the curvature (v) at a pxticolar section, or the rotation (0)at a joint, or the displacement (A) at a selected point. The ductility ratios

'

However, very high grades of concrete are undesirable, as the they have lower ultimate compressive strains [refer Fig. 2.71.

774 REINFORCED CONCRETE OESIGN

moment M

load-displacement response as shown scl~ernaticallyin Fig. IG.l(c). This ductility is achieved by ensuring dactilc membcx section responses (as indicated by the M-rp relation in Fig. l6.l(a)), so that an adequate membcr of plastic hinges [refer Section9.71 would develop at appropriate locatiom under cxtrerne lateral SC~SINC forces.

ldeallsed actual

16.2.2 Energy Dissipation by Ductile Behaviour flexural member

*"

strains

curvature q

*"

(a) curvature ductility for an under-reinforced section

t (b) member behav~our

v.

> 8, A

displacement A (C)

struclure behaviour Fig. 16.1 Measures of ductility

The problem becomcs more complex when it cmnes to defining a ductility measure for an entire structure. In general, a reinforced concrete ductile structure will have a load-displacement response as shown schematically in Fig. 16.l(c). This ductility is achicved by ensuring ductile member section responses (as indicated by the M-p relation in Fig. 16.l(a)), so that an adequate member of plastic hrnges [refer

Under seismic forces, structures are subject to several cyclcs of reversed cyclic loading. If the structure, modelled (for sirnl~licity)as a single degree-of-freedom system, were to behave in a h e a r elastic manner under reversed cyclic loading, i t will exhibit a linear load-displacement behaviour as shown in Fig. 16.2(a). T h e shaded area under the curve denotes the potential energy stored in the structure at the maximum displace~nentposition; this gets released and converted to kinetic energy a s the structure returns to its zero-load position. However, if the structure responds in an elastoplastic (ductile) manner, developing fully plastic behavionr at a load level F. (less than F shown in Fig. 16.2a), the11 the load-displacement behaviour is as shown in Fig. 16.2(b). In this casc, the maximum deflection A' is grealer than that (A) obtained in Fig. 16.2(a) for claslic behaviour. Furthenno~e,when thc structure ~ t u n to ~ its s zero-load position, the actual energy which gets converted to kinetic energy is limited to the triangular area crle in Fig. 16.2(b). The re~nainderof the input energy (given by area abccl) gets rli.rsipotcdt by the plastic hinge. In summary, under seismic loadiogs, for a given mergy input, elastoplastic response differs from elastic response in tbc iollowi~lgways:

.

the energy gets dissipated; the induced force is less; and lhc maximum deflection is more.

It may be noted that the actual behaviour of reinforced concrcte is different from the idealised behaviour shown in Fig. 16.2(b). As indicated schematically in Ag. 16.2(c), the hysterisis bchaviour of reinforced concrete is characterised by 'rounding' and 'pi~~ching' of the loops, which is associated with the Bouschi~~ger effect* in steel, and sc$'tiircs,~~Iegradationin concrete (due to repeated ope11i11gand closing of cracks and bar slip at anchorage zoncs) [Ref. 16.111. This results in the areas within successive loops becol~llgsmaller.

-.

gets canvcrted into heat and other folrns of nonrecoverable energy. steel is sub,jected to reversed cyclic loading, it is found that the yield When stmlgth obtailled in the icloading or revet.sed direction is substantinily less rbnn llle initial yield slrengti~:this is Bo\vn ss rlie Bosschirtgcr. c//ecr [refer Fig 2.191.

'

776

RE!NFORCED CONCRETE

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANT DESIGN

DESiGN

777

16.2.3 Flexural Yielding in Frames and Walls As reinforced concrete is relatively less ductile i n compression and shear, dissipation of seisrnic cnergy is best achieved by flexural yielding. Hence, weakness in potential energy converted to kinetic energy in one cycle

A

compression and shear, i n relation to flexure, should be avoided. I n a structure composed of ductile moment-resisting frames andlor shear (flexural) walls, the desired inelastic (ductile) response is devclopcd by the formation of plastic hinges (flexural yielding) in the members, as shown in Fig. 16.3.

displacemenl

(a) elastic response

P.E. converledto K.E.

in one cycle isplacement

7

(i) seismic loads (equivalent static)

(iii) hinges in beams

(ii) hinges in columns

(a) ductile frame

(b) elastoplastic response

load

?

/first

I

loop

shear (flexural) wall

isplacement

_plastic hinge (c) hysteris behaviour of reinforced concrete

Fig. 16.2 Load-displacement behaviour under reversed cyclic loading

rn

(b) ductile wall Fig. 16.3 Formation of plastic hinges in a ductile structure

778 REINFORCED CONCRETE DESIGN

SPECIAL

In the case of ductile frames, plastic hinges may form in the beams or in the columns, as shown in Fig. 16.3(a). It is desirable to desien thl: frame such that the plastic hinges form in tllebeams [Fig. 16.3(a)(iii)], and not the columns, because: e plastic hinges in beams have larger rotation capacities than in columns; mechanisms involving beam hinges have larger energy-absorptive capacity on account of the larger number of beam hinges (with large rotation capacities) possible; e eventual collapse of a beam generally results in a localised failure, wlicreas collapse of a colmnll niay lead to a 'global' failure; and * colu~nnsare more difficult to straighten and repair than beams, in the event of residual deformation and damage.

-

16.3 MAJOR DESIGN CONSIDERATIONS

16.3.1 General Design Objectives The objective of the special design and detailing provisions in IS I3920 [Ref. 16.31 is to ensure adequate lougl~nessand ductility (with ability to undergo large inelastic reversible deformations) for individual members such as beams, colu~nnsand walls and their connections, and to prevent other non-ductile types of failure.

PROVISIONS FOR EARTHQUAKE-RESISTANTDESIGN 779

resulting in thedevelopment of reversible plastic hinges at various locations in the ductile frames and walls. The structural system should be so designed as lo ensure that the formation of plastic hinges at suitable locations niay, at worst, result in the failurc of individual elements, but will not lead to instability or progressive collapse. This calls for building-in redundancy into the structural system. Redundancy assists in the development of alternative load paths, tliercby helping redistribution of forces, dissipation of encrgy and avoidance of progressive collapse. ""

suggests that the anticipated drift due to seismic forces may be taken as three limes the lateral deflection obtai~icd from the usual elastic analysis under equivalent factored static loads. [This factor is intended to account for the effects of material and geometric nonlinearities, as well as additional amplificatio~ldue to dynamic effects]. The inter-storey drift is to be limited to 0.004 times thc storey height (under the specified seismic forces) as per IS code [Ref. 16.1]. The effect of drift on the vertical load-carrying capacity of the lateral load resisting system should also be taken into account in the analysis. 16.3.3 Materials Reinforcing S t e e l

Some of the main design considerations in providing ductility include: using a low tensile steel ratio (with rclatively low grade stecl) andlor using compression steel; o providing adcquate stii~upsto ensure that shear failure docs not precedc flexural failure; confining concrete and compression steel by closely spaced hoops or spirals; and e proper detailing with regard to connections, anchorage, splicing, iiunimum reinforcement, etc. Furthermore, continuity in construction and redundancy in structural framing are desirable for the devclopnient of more illelastic response, and thereby more moment redistribution and energy dissipation at sevcral plastic hinges. Earthquake is often followed by fire and hence fire resistance should also be a major consideration in building construction in seismic regions.

16.3.2 Requirements of Stability a n d Stiffness Under a sevcre earthquake, it is expected that io a strxture designed to resist seismic forces in a ductile manner, large lateral deformations and oscillations will be induced,

As mentioned earlier, ductility calls for the use of relatively low grades of steel. Lower grade steel has clearly dcfined and longer yield plateau, and lience the plastic hinges formed will have larger rotation capacities, leading to greater energy dissipation. Similarly, locations of potential plastic hinges sltould nor have roo much over-strength, i.e., wengtli morc than the required design strength. Over-strcngth will result in the section not yieldiug, as intended, at the expected lateral Load levels. This may result in adjoining elements andor foundations being subjected to loads larger than the d e s i g ~loads, ~ with consequent damage. 111 other wonls, the acrual yield strength of the steel used should nor be rnnrkrdly higher than the yield strcllgth specified and used in design computations. Furthermore, yield strength, far in cxcess of that specified, may lead to cnccssivc shear and bond stresses, as the plastic moment is developed. Another point to note is that, the lower the grade of steel, the higllcr- is the ratio of tllc altirnnte tensile strength (L,) to the yield strength V), [refer Section 2.14.21. A high ratio off& is desirable, as it results in an increased length of plastic hinge (along the member axis), and thercby an increased plastic i-otarion capacity. For these reasons, mild steel (Fe 250) is best suited for use as flexural reinforce~nentin earthquake-rcsistal~tdesign. However, its use will necessitate larger sections of flexural members. Hence, the code [Ref. 16.31 pennits the use of the higher grade Fe 415 (which is most commonly used in practice), but prohibits the use of grades higher than Fe 415. ,

780 REINFORCED CONCRETE DESIGN

Concrete With regard to the gradc of concrete, the code [Ref. 16.31 1i1& the minimum grade of concrete to M 20 f i r all buildings which are more than 3 storeys in height). It may be noted that very high strength concrete is also undesirable because higher compressive strength is associated with lower ultimate compressive strain (E,,,) [refer Sectio112.8.2, Fig. 2.71 - which adversely affects ductility. Likewise, low density concretc is undesirable because of its relatively poor perlormance under reversed cyclic loading. The ACI and Canadian codes [Ref. 16.4, 16.61 limits the nlaximum cylinder strength of low density concrete for use in earthqoake-resistant design to 30 MPa.

16.3.4 F o u n d a t i o n s It is important to ensure that the foundation of a structure does not fail prior to the possible failure of thc supcrstrocture. As plastic deformations are pcrmitted to occur at suitable locations in the superstmcturc under a scverc earthquake, the maxinlum seismic forccs transmitted to the foundation will be governed by the lateral loads at which actual yielding takes placc in the skuctural elcmcnts transferring the lateral loads to the foundation. The ultimate moment, correspontli~lgto 'nctual yielding' at a section is obtained as its chrrmcteristic (nominal) momcnl capacityt, i.e., without applying partial safety factors (i.e., with y, = y, = 1.0). The corresponding moments, shear forces and axial forces transferred from the fiames and walls to the foundation system (under conditions of 'actual yielding') should be resisted by the foundation system with the usual margin of safety (i.e., with ?: = 1.5 and y, = 1.15) in order to cnsure a combination of a relatively stronger foundation and wcaker superstructure. Although such a recommendation is yet to be incorporated in thc IS codes [Ref. 16.1-16.31, it is in vogue in several international codes (such as Rcf. 16.4, 16.7). Such a dcsign concept is ~~ecessnry to provide for ductile behaviour or the superstructure witlmut serious damage to the foundaliot~.

.

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANTDESIGN 781 To avoid sudden brittle failure of a beam (when the cracking moment of the section is reached) a. minimum reinforcement ratio, p,,,., = 0 . 2 4 z 1 f, , must be provided at both the top and bottom for the entire length of the member, with at least two bars placed at each face. Flexural members of lateral force resisting ductile fiames are assumed to yield at the design earthquake. To ensure pmper development of reversible plastic hinges near continuous supports (beam-column connections) where they usually develop in such members, the 'positive' moment reinforcement at a joint face must not be less than half the 'negative' moment reinforcement at that joint face: r the top and bottom steel at any section along the length of the member should not be less than ane-fourth of the 'negative' moment reinfoxcement at the joint face on either side; * both top and bottom bars must be taker1 through the column and made continuous wherever possible, in case of an interior joint. In other cases, they must be'extcnded to the far face of the confined colrlrnn core and provided an anchorage length of Ld + 10 g, where 4 is the development length of the bars (diameter $) in tension [Fig. 16.41: and * Not more than 50 percent of the bars shall be spliced at one section. Because of the possibility of spalling of the concrcte shell (covcr) under large reversed strains, lap splices of flexural reinforcement are not permitted in and near possible plastic hinge locations. If welded splices or mechanical connections are used, it must be ensured that not more than 50 percent of the bars are spliced in the region of potential plastic hinging. * The provisions for redistribntion of tnonlents (See Section 9.7.3) shall be used only for vertical load moments and not for lateral load moments.

1 4 3 . 5 Flexural Members in Ductile F r a m e s

The code i~conunendations[Ref. 16.31 for design and dctailing of flexural nlembers in earthqoake-resislaot design nre as follows: 0 To qualify as "flexural membels", the factored aninl strcss under earthquake loading should not exceed 0.1 fck. Further, the overall depth D should not exceed one-fourth oC the clear spau (to limit shear deformations) and rhc width b sllould not be lcss than 200 nun, with a W D ratio of marc than 0.3 (to avoid lateral instability and provide for improved torsional resistaocc). s To ensure signiiicmt ductile behaviour even under reversals of displacements in the inelastic range, to avoid congestion of steel, ~ I K I to limit the shear stresses in beams, the tensile reinforcement ratio p, is limitcd to 0.025 i.e., p,, ,,, = 2.5. -

' ~lte;nstirrl~,this

rnkcn (conservatively) as 1.4 timer the factored moment of resistance (M,,") - as recotnmended by the code [Ref. 16.31. for estimating plastic manlent capcities in the cdculnlion of dcsign shear forces. may be

Flg. 16.4 Anchorage of beam bars at an external joint

SPECIAL PROVISIONS FOR E A RTH Q UA KE - RE S IS T A NT

782 REINFORCED CONCRETE DESIGN e

When lap splices are provided (at regions other than plastic hinging regions), transverse reinforcement for confining concrete and to support longitudinal bars. in the form of closed stirrups or 'hoops'.(with a 135' hook and 10 I$ 2 75 mm extension) should be provided over the entire splice length, at a spacing not exceeding 150 nun [Fig. 16.5]. The bar extensions lnust provide for possible shifts in the inflection points, which n~ayoccurunder the combined effccts of gravity and seismic loadings. During an earthquake, a structure should be capable of undergoing extensive inelastic deformation (through ductile behaviour) without a significant loss in strength. Yielding softens the structure, which effectively increases its time period and reduces the earthquake force. Damping also increases significantly in the inelastic range of response and this further helps to improve the earthquake response. For these desirable effects to take place, it should be ensured that none of the brittle modes of failure (particularly, shear failure) should occur before ductile flexural failure. Hence, the shear design philosophy in an earthquakc resistant structure differs significantly from that in an ordinary structure [Ref, 16.20, 16.241. Due to extensive cracking i n the zones of liigh shear, it is desirable to con~pletelyignore the shear strength of concrete (2,) and to design the stirrups to resist the entire shear.

the effccts of factored gavity loads and sway in either direction, are indicated in Fig. 16.6(c). The maxim~m~ design shear forces (V,,) at the s ~ p p o faces f (left or right) are accordingly obtained as: (16.2a) V , , , , = 0.5w,,LZ+ 1.4 (M,i,lep +MI&i8er )/l.

where C, is the c l e k span, and I",, = 1.2(w,

+ wLL)

(16.2~)

assuming that the gravity loads (dead loads WD, and live loads distributed.

2%

lvd

e

beam

(b) design shear forces in beam (sway to right)

(c) design shear forces in beam (sway to lefl)

-

' The factor of 1.4 sllecified by the code [Ref, 16.31is intended to account for the condition of 'actual yielding' (involving clmracferisfic values of material strengths) as well as increased tensile strength due lo possible strain hardening, and also some margin of safety.

are uniformly

(a) loading on

Flg. 16.5 Lap splice in a flexural member

In cnrth~nakeresistant structure, the design - shear force will be the lamer of 1. Shear force as obtained from analysis for given load con~binations,and 2. Actual shear force likely to develop in a member after flexural failure has taken place. According to the code (CI. 6.3.3; IS 13920: 1993), the web reinforcement in the farm of vertical stirrups shall be provided so as to develop the vertical shear duc to formation of the plastic hinges at both ends of the beam plus the factored gravity load on the span. To ensure that a shear failure does not precede the full development of plastic hinges in a beam, the design shear forces in the member should be suitably overestimated, considering plastic moment capacities' of 1.4 M,,# at the beam cnds, as shown in Fig. 16.6(b). The component shear force diagrams, including

DESIGN 783

Fig. 16.6 Calculation of design shear forces in beams

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANTDESIGN 785

784 HEINFORCED CONCRETE DESIGN

! ti I

: ,

s

o

Because of thc alternating direction of the shear force due to seismic effects, the direction of the associated diagonal tensile stress also alternates, as shown in Fig. 16.7(a). For this mason, ir~clined0413 (which are effective only against shear in one direction) are not allowed as effective shear reinforcement. Web reinforcement for seismic design lnust be in the form of closed stirrups, called hoops, placed yerjper~diciicrrlarto the lo~zpitndi,talwinforce,rre,tt and mtlst be provided throughout the length of the member. These hoops should havc a minimum diametcr 9,of 8 mm in bcams with a clear span exceeding 5,,1 (6in shorter beams). Thc ficc ends of the hoops should be bcnt at 135" with a minimum bar exteusion of 10 $, ( b y not < 75 mm) [Fig. 16.7(b)l, so that the ends areadequately anchored in the core of the conirete.

,inclined bars effective -. ........ . . . .. . . . . . . . hoops for vertical shear

nc lned oars

- ..

-. ......

8 times the diameter of the smallest longitudinal bar, with the first hoop located at a distance not exceeding 50 mm from the column face. Elsewhere, in the beam, the spacing of hoops should not exceed d12, as shown in Fig. 16.7(c). 16.3.6 Columns and Frame Members Subject to Bending and Axlal h a d e

.

-

-,

.

. Fig. 16.7 Type of web reinforcement for reversed shear condilion a

Thehoopuerve the adrlitio~~al pmposcs of confining the concrete and preventing buckling of the longitudinal bars, particularly ncar the beam-column joints, where reversible plastic hinges are expected to dcvelop atld where the concrete cover is liable to spa11 uff aner a few cycles of inelastic rotations. Thc code [Ref. 16.31 specifies a closes spacing of hoops over a length equal to twice the effcctivc depth (i.e., 2d) from the face of the column. The hoop spacing should [lot exceed dl4 or

'

Members in this category are those having a factored axial stress which is greater than 0.1& mder the effect of seismic forces. Further, the minimum dimension of the mcmber should not be less than 200 m , with the ratio of the shortest crosssectional dimension to the perpendicular dimension preferably not being less than 0.4. Howcver, in frames having beams with centre to centrc span exceeding 5 m or columns with unsupported length exceeding 4 m, the shortest dimension should not be less than 3 0 0 m [Ref. 16.31. T o ensure that the combined flexural resistance of the columns is greater than that of the beams at the beam-column joint (so that the plastic hinges form at the beam ends, rather than the column ends), it is necessary to design the collllnn sectiol~for a suitably higher moment. Although the IS code [Ref. 16.31 does not make any specific recommendation in this regard, the ACI and Canadian codes [Ref. 16.4, 16.71 mcomnel~dthat the sum of the factored moment resistances of the columns framing into the joint be at least 1.1 times the sum of the charucteristic momcnt resistances (i.e., = y, = 1.0) of the beamsr framing into the joint [Fig. 16.8(a)]. Lap splices are not permitted near the ends of the column wherc spalling of the concrete shell is likely to occur. Lap splices (suitably designed as tension splices), however, are permitted in the cenWal half of the member ler~gth. Hoops should be provided over the entire splice length at a spacing not exceeding 150 mm (centreto-centre). Not more than 50 percent of the bars should be spliced at any section. The design shear f o x e in a column should be taken as the larger of (1) the shear forcc doe to the factored loads and (2) the shear force in the column due to the developmeut of the plastic moments (suitably enhanced, as in Eq. 16.2) in the beans framing into the column, given approximately by [Ref. 16.31: (10.3) V,,=1.4(~,,,,,1 + M ,, ,,)lh,, b2 are the factored moments of resistance (of opposite sign) where MI,,, and of beam ends '1' and '2' fiaming into the colulnn from opposite faccs, and II,, is the storey height [refer Fig. 16.8(b)l. Unless a larger amount of transverse reinforcement is required from shear strength considerations, special confining reinforcen~entshould be provided as given below. Special confining reinforcement must be provided over a length I, from each joint face.(high moment regions), and on both sides of any section, where flexural yielding may occur under seismic forces [Fig. 16.9(a)]. The length l,, should not bc less than (a) the larger lateral dimension of the member at the section where yielding may occur, (b) 116 of the clear span (height) of the member, and (c) 450 mm. The spacing of hoops used as special confining

The effects of the slab reinforce~nentwithin a distance of three times the slab thickness on either side of the beam should bc included in calculating tile bean mornenl capaciry.

!

3

t: * ,,

:

786 REINFORCED CONCRETE DESIGN

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANT DESIGN 787

:,!

,! reinforcenlent sllould not exceed 114 o f the minimu~imemberdimension, but need be less than 75 nun, or more than 100 mm. The area of cross-section (A,,) of the bar to be used as special confining reinforcement should be taken as: for circular hoopslspiral A ,

2

(16.4)

(a) detailing of hoops in column (at and near joint)

---

REINFORCEMENT

where s s pitch of spiral or spacing o f hoops; D, diameter of core, measured to the outside o f the spiral or hoop; D , longer dimension of the rectangular hoop, measured to its outer face - ,lot to exceed 300 mm; A, gross area of the column section: and Ai area of the concrete core (contained within the outcr dimellsion of the hooplspiral).

1 1

swa

SPECIALCONFINING REINFORCEMENT

(b) detailing at column-footing interface

Me,, -r factored ultimate moment caoacitv of .

column end

--* Me,,

.

+ characteristic Ultimate moment capacity of beam end = 1.4 MUe,*,



(a) moment resistance requirement

I

(c) special conlining reinforcement requirement for columns under discontinued walls

relatively stifl columns (attracting large seismic forces)

(d) columns with varying stiffness

+

(b) shear resistance requirement

W

Fig. 16.8 Column resistance requirements

Fiy. 16.9 Special confining reinforcement

SPECIAL PROVlSlONS FOR EARTHQUAKE-RESISTANTDESIGN 789

'88 REINFORCED CONCRETE DESIGN

I

When a column terminates into a footing or mat, the 'special confining reinfotzentent' should extend at least 300 nun into the footing or mat, to account for possible development of plastic hinges at the base of a building [Pig. 16.9(b)]. Such detailing should also be provided in columns supporting discontinued stiff members (such as walls or trusses) for the full height of. the CO~LUIIIIas shown in Fig. 16.9(c). Pmvisioll of special confining reinforceme% over the full height of the column is also required in cases where there is a significant variation of stiffness along the height of the column - as when mezzanine floorsllofts are provided locally [Pig. 16.9(d)l or due to in-filled masonry walls not extending fully over the panel.

.

16.3.7 Joints in Ductile Frames

* Beam-column joints in ductile frames must have adequate shcar strength and

0

ductility to facilitate the development of large inelastic reversible rotations, in thc event of a severe earthquake. Tests have indicated that the shear strength of joint7 is dependent primarily on the grade of concrctc and is not sensitive to the amount of shear reinforcement [Ref. 16.41; hence, it is desirable to use high strength concrete in the joint regions, and to achieve good comlxction of this concrete [Ref. 16.121. Thc special confinixg ,rinfo,-cement (hoops) provided near the c o l u ~ mends should be extended througk the joint ns well [see Fig. 16.9(a)J. However, when the joint is 'externally confined', this reinforcement may be reduced to one-half of that required at the end of the column, with the maximlm spacing limited to 150 mm (Code CI. 8.2). A joint is said to be 'externally confined' if beams frame into all the vertical faces of the joint, and if each beam width is at least threefourths of the column width at the joint [Ref. 16.3 &16.12]. Development length requirements of the flexural reinforcement within the joint [refer Fig. 16.41 arc particularly important. The joint zone is an area of high concentration of beam bars, column bars and hoops. Extreme care is needed in detailing tlle reinforcement at the bcam-column connection in order to provide for proper stress traosfcr, and to avoid congestion and placing diiliculties for both reinforcement and concretc many structurd failulcs under scisrnic loading can be traced to pooj detailing of beatn-column joints. The reinforcements detailed in this chapter pertain to mo~~olitbic concrete consuuction In precast constroction, subject to seismic loadine, the most critical location is the beam-column cortncction. However, it has been sllown that by careful detailing, ductile beam-column connections (having adequate strength, stiffness, ductility m d energy-dissipating capacity) can be made in precast concrete constmction as well [Ref. 16.19].

16.3.8 Shear Walls (Flexural Walls) e

Ductile 'shear walls' (more appropriately calledflexuml wrrlls), which form part of tlle lateral load resisting system, are vertical members cantilevering vertically from the foundation, dcsigned to resist lateral forccs in its own plane, and are subjected to bending moment, shenr and axial load. Unlike a beam, a wall is

e

.

.

relatively thin and deep, and is subjected to substantial axial forces. T h e wall must be designed as an axially loaded beam, capable of forming reversible plastic hinges (usually at the base', as shown itt Fig. 16.3(b)) with sufficient rotation capacity. The code [Ref. 16.31 recommends that the thickness of any part Of the wall should preferably be not less than 150 mm. Walls that are thin are susceptible to instability (buckling) at regions of high compressive strain. Stability of the compression zones can be improved by local thickening of the wall or by providing flanges m. cross walls (which is convenient at such locations as lift cores). Flanged walls also have higher bending resistance and ductility. T h c code [Ref. 16.31 restricts the effective flange width of flanged walls to (a) half the distance to an adjacent shear wall web, and (b) one-tenth of the total wall height. The wall should be reinforced with uniformly distributed reinforcement in both vertical alld horizontal directions, with a minimum reinforcement ratio o f 0.0025 of the gross section in each direction. The bar diameter should not exceed onetenth the wall thickness, and the bar spacing in either direction should notexceed (a) 115 of the horizontal length of wall, (b) thrice the wall (web) thichess, and (c) 450 mm. The distributed reinforcement provides the shear resistance, controls the cracking, inhibits local breakdown in the event of severe cracking during an earthquake, and also resists shrinkage and temperature stresses. The vertical reinforcement, comprising both the distributed reinforcement and concentratcd reinforcement near wall ends (sec below), should be designed for the required flexural and axial load resistance. In walls which do not have flanges ('boundary elements'), concentrated vertical reinforcement should be provided towards each end face of the wall, in addition to the uniformly distributed steel. A tninitnum of 4 nos 12 mm 4 bars arranged in at least two layers should be provided near each end face of the wall [Ref. 16.31. The concentrated vertical flexural reinforcement near the ends of the wall must be tied together by transverse ties, as in a column, to provide confinement of t h e concrete, aad to ensure yielding without buckling of the compression bars when a plastic hinge is formed. Where the extreme fibre compressive stress in the wall exceeds 0.2 f,, boundmy elements should be provided along the vertical boundaries of walls. These are portions along the wall edges that are strengthened by longitudinal and transverse reinforcement, and may have the same or larger thickness as that of the wall web. To prevent a premature brittle shear failure of the wall before the development of its full plastic resistance in bending, it is desirable to design the shear resistance of the wall for an overestimated shear force, as in the case of the column. Because of possible severe sl~aarcracking under reversed cyclic loading, the shear carried by concrete in thc plastic hinge region is neglected. For othcr details regarding the design of boundary elements, coupled shear walls, walls with openings, etc., reference should be made to Ref. 16.3. -

p~

--

'Locations of ablupt changes in the strength and stiffness of the lateral load resistmg system are also potential zones of flexural yielding in ductile walls.

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANTDESIGN 791

16.3.9 lnfill Frames Generally, in the analysis of multi-storey buildings, the contribution of masonry infill walls is ignored, and the frame analysis is based on the bare RC frame. The mass of the masonry infill is considered, but the stiffness and strength contrib~~lions of the masonry infill are neglected. However, the infill frame has some significant effccts under lateral loading that merit consideration [Ref. 16.21, 16.221: Infills alter the behaviour of buildings from predominantly frame action to predominantly shear action [Pig. 16.101. Also, the infills are capable of resisting the applied lateral seismic forces through axial compression along the diagonal; there is no tensile resistance capability in the other diagonal, but the cracking induced in the masonry on account of this serves to dissipate energy. a The neglect of infill contribution results in a significant under-estimation of the lateral stiffness of the structure, and thereby can result in an under-estimation of the seismic forces. Infills may also significantly modify the position of centre of rigidity and consequently can affect the behaviour in torsion. In regular multi-storey buildings, in gencral, the neglect of infill frame action results in a conservative estimation of bendillg moments in columns and bcams (except when 'soft storey' is provided, as shown in Fig. 16.11).

separation of frame occurs from the infill at early stages. The panels are in contact with the frame only at thc compression corners, and this contact is strengthened under increased loading, with high stress concenttations near the comers. The diagolral part of the infill acts as a comprwsive diagonal strut and is effective in resistillg lateral loads. As the tensile con~ersare subjected to very small stress, the tensile diagonal region is not really eflective in resisting lateral loads. Since the infills act as diagonal struts, an infill wall can be replaced by an equivalent strut in the analysis model.

16.3.10 sift Storey The soft storey concept is related to a discontinuity in tlle stiffness of building. According to IS 1893: 2002 a soft storey is one in which the sum of the lateial translational stiffness is less than 70% of that in the storey above or less than 80% of the average latenl translational stiffness of three stories above. In modern multistorey construction, such soft stories are com~nonlyencountered in the ground storey. Owing to high cost of land and small sizes of plots, parKng is often accommodated in the mound storey area of the building itself. Franle bays of tile ground storey are not infilled with masonry walls, as in the case of upper stories. Usually, all panels are left open for parking. Thc suilden discontinuity in stiffness and mass at the lowermost storev .. leads to thc following effects that makc soft storey construction , ,(soft storey) particularly dangerous. The stiffness discontinuity leads to severe stress concentratio~lsat the soft storey comers, accompnied by large plastic deformations [Rg. 16.1 l(a)l. Most of the deformadon energy is dissipated by the soft storey columns, and this leads to major overstressing of these elements; onset of plastic hinges lnay transform the soffstorey into a mechanism resulting in collapse. Such a collapse .~~ could also turn out to bc more catastrophic.

.

~~~

(a) Bare frame:predominant frame action

(b)

lnfiilsdframe: predominant Shear action

Fig. 16.10 Behaviour of infill frame [Ref. 16.231

Thus, the barc RC frame of the building considered in design is inconsistent with actual behaviour. In general, however, the infills are expected to significantly mduce thc demand on the RC frame members. Numcrous cases dre cited in the technical literature where brick walls acting together with RC elements have saved buildings from collapse during earthquake. Several techniques have bcen proposcd to evaluate the allowable horizontal force of an infilled frame subject to in-plane bending and axial force. The simplest procedure is to model thc masonry infill by means of an equivalent compressive diagonal strut [Ref. 16.211. At the tensile corners of the non-integral infill walls,

(a)Open ground slorey

(b) Bare frame

Flg. 16.11 Lateral load responses of open ground storey fratre (with infillsabove ground storey) and bare frame

SPECIAL PROVISIONS FOR EARTHQUAKE-RESISTANT DESIGN 793

992 REINFORCED CONCRETE DESIGN

For designing a 'soft storey' building, dynamic analysis should bc carried out including the strength and stiffness effects of infills and inelastic deformation in the members. Alternatively, the codc suggests the following design criteria based on a conventional earthquake analysis, neglecting the effect of infill wails in other storiest. The colynns and beams of the soft storey are to be dcsigned for 2.5 times the storey shears and moments calculatcd under scismic Load (which ignoms thc infill framc effect). 16.3.11 Performance Limit S t a t e s In several countries, seismic design is in the process of fundamcotal change. One important reason for the cllange is that although code-designed buildings pel-fonned well (in countries such as USA) in recent earthquakes from a life-safety perspective, the level of damage to structures, econonuc loss due to non-usage of buildings and costs of repair were unexpectedly high. Conventional methods of seisnlic design have the objectives to provide for life safety (through strength and ductility) and damage control (through serviceability - drift li ts). The design criteria are defincd by limits on stresses and member forces calculated from prcscribcd levcls of applied lateral shear force. Performance-based design philosophy involves dcsign criteria that are cxlxessed in tcrms of achieving stated performance objcctivcs whcn thc structure is subjected to stated levels of seismic hazard [Ref 16.251. The perfonnnnce targets may be a level of stress, a load, a displpccmncnt, a limit statc or a targct damagc state not to be exceeded. Required perforn~ancecriteria for a seismic hnzard are 'safety', 'restorability' and 'usability'. Safety refers to protection of human life. Restorability refers to structual integrity. Usability refers to function and habitability. 16.4 CLOSURE Thc purpose of this chapter is to explain the background to the seismic design provisions of the IS code and related international codes. A detailed discussion of seis~nicanalysis and design of reinforced concrete structures is beyond the scope of tius book. Rapid advanccs arc being made in this area, and recent publications (for example; Refs. 16.13-16.18) may be consulted for more details.

16.5 What are the objectives behind the special detailing provisions in I S 13920? 16.6 Differentiate between the terms strength, stifltess and stability as applied to a remforced concrete structure. is it desirable to design for the formation ofplasric hinges in beams rather 16 7 Whv -~ -~ , than columns in carthquakc-resistant design? 16.8 Is it desirable to have (a) high strength steel (b) high strength concrete in earthquake-resistant design of reinforced concretc structures? Justify your answers. 16.9 Suggest a design procedure for ensuing that the foundation is stronger than the superstructure in carthquakc-resistant design. 16.10 What are the limits placed on tensile reinforcement ratios in beams in earthquake-resistant design? Why are such limits enforced? 16.11 How are the design shear forces estimated in the beams of ductile frames? 16.12 Why are inclined stinups and bent-up bars unsuitable as shear reinforcement in earthquake-resistant design? 16.13 What is mcant by . special confining reinforcement in columns of ductile frames? 16.14 What are the design requirements of beam-column joints in eartbquakeresistant design? 16.15 Explain the differences between an ordinav wall and a shear wall in a reinforced concrete tall building, with regard to function, loading and design. 16.15 What are the main design requirements of ductile shear (flexural) walls in earthquake-resistant design? 16.17 What is the effect of ignoring the contribution of masonry infill in the lateral load analysis of a multi-storey frame? 16.18 In what manner is the behaviour of a 'soft storey' construction likely to be different from a regular construction in the event of an earthquake?

-

~

REFERENCES IS 1893 (Part 1): 2002 - Cviteria for Earthquuke Design of ~trhcturesPnrrI: General Provisions and Buildinjis (Fifth revision), Bureau of Indian Standards, New Delhi, 2002. 16.2 1S 4326 : 1993 - Code of Practice for Earthquake Resistant Design and Construction of Buildings, Bureau of Indian Standards, New Delhi, 1993 (reaffirmed 1998). 16.3 IS 13920 : 1993 - Ductile Detailing of Reinforced Concrete Structures Subjected to Seismic Forces - Code of Pracrice, Bureau of Indian Standards, New Delhi, 1993. 16.4 ACI Standard 318-95, Building Code Requiren~enrsfor Srrucrural Concrere and Commentary (ACI 318R-95). Am. Conc. Institute, Detroit, Michigan, 1995. 16.5 National Building Code of Canada 1995, Part 4: Strucrural Design, National Research Council of Canada, Ottawa, 1995. 16.6 CSA Standard A23.3 - 94 - Design of Concrete Stri~!ru,rs, Canadian Standards Association, Rexdale, Ontario, 1994.

16.1

~

REVIEW QUESTIONS 16.1 What are the objectives of earthquake-resistant design of reinforced concrete structures? 16.2 *hat is meant by ductility? Give a qualitative description and also describe briefly the qualitative measures of ductility in reinforced concrete. 16.3 What are the measures one can take for improving the ductility of a reinforced concrete structure? 16.4 What are the advantagesldisadvantagesof elastoplastic bchnviour over elastic behaviour in structures subjected to severe earthquakes? 'The desigu criteria have beetr.newly introduced in the recent (2002) revision of IS 1893. If these arc applied to existing buildings, it will be seen that a mnjority of such buildings will be found deficient in terns of earthquake resistanl design

~

794 REINFORCED CONCRETE DESIGN

16.7

SEAOC, Recommended Lateral Force Requirements and Cornnrentajy, Seismology Committee, Structural Engineer's Association of California, San Francisco, 1980. 1 6 8 - Explanatofy Handbook or Codes for Earthquake Engineering, Special Publication SP 22, Bui'cau of Indian Standards, New Delhi, 1982. 16.9 Ncwmark,N.M. and Roscnblcuth,E., Fundanrentals of Earlhqunke Enginee~ing.Prentice-Hall, Englewood, Cliffs, N.J., 1971. 16.10 Clough. R.W. and Penzien, J., Dyriamic.~of Structures, Second edition, McGraw-Hill lntemational edition, 1993. 16.11 Park, R. and Paulay, T., Reinforced Concrete Structures, John Wiley & Sons, Inc., New York, 1975. 16.12 ACI-ASCE Committee 352, Recoramendations for Design of Bearn-Colurnn Joints in Monolithic Reinforced Concrete Structures, (AC1352 R-76, Reaffirmed 1981), Am. Conc. Institute, Detroit, 1976. 16.13 A~nold,C. and Reitherman, R., Building Configurntion and Seismic Design, John Wiley & Sons, Inc., New York, 1982. 16.14 Applied Technology Council, Tentative Provisions for the D c v e l o p e r ~ tof Seismic Regulatiorrs for.Buifdings, ATC 3-06, National Bureau of Standards, Special Publication 510, U.S. Government P~.intingOffice, Washington, D.C., -~ 1978. 16.15 Dowrick, D.J., Earthquake Resistant Design, John Wiley & Sons, Chichester, U.K., 1977. 16.16 - Reinforced Concrete Srructwes in Seisrnic Zones, ACI Publicatio~lSP-53, A m Conc. Institute, Detroit, 1977. 16.17 - Reinforced Concrete Str.uctures Subjected to Wind and Earthquake Forces, ACI Publication SP-63, AmConc. Institute, Detroit, ,1980. 16.18 -Earthquake ESfects on Reirrforced Concrete Structure, ACI Publication SP84. Am. Conc. Institute, Detroit, 1985. 16.19 Pillai S.U., and Kirk, D.W.. Ductile Bearn-Colurtin Corinecrion in Precast Concrete, ACI Journal, Vol. 78, Nov-Dec 1981, pp 480487. 16.20 Murty C.V.R., and Jain S.K., A Review of IS 1893-1984: Plnvisior~son Seismic Design of Buildirrgs, The Indian Concrete Journal, November 1994, pp 619-629. 16.21 Scadat A.S., Approxirnate Methods in Structural Seisrnic Design, E & FN Spon, London, 1996. 16.22 Mallick D.V., and Sevem R.T., The Behaviou?. of Ir~@lled Frames, Proceedings of Institute of Civil Engineers, Vol. 38, Dec 1967, pp 639-656. 16.23 - Rhuj, India Earthquake of January 26, 2001: Reconnaissance Reyo,% Supplement A to Volume 18 Earthquake Spectra, July 2002. 16.24 Medhekar MS.. Gehad E.R.. and Jail1 S.K., Shear Reinforcement for.Aseismic Design of Flexural Members, The Indian Concrete Journal, June 1992, pp 3 19-324. 16.25 Ghobarah A., Pe$oi?~lance Based Design in Earthquake Engineering: State of Development, Engineering Structures, Vo1.23,2001, pp 878-884

17.1 DESIGN FOR SHEAR BY COMPRESSION FIELD THEORY 17.1.1 Introduction

~

In the traditional method given in Chapter 6, the transvcrsc reinforcement for shear is designed separately and added on to the reinforcement designed for flexure (with ~ Influence of shear on longitudinal reinforcement axial load if any), and f o torsion. requirements is taken care of by detailing This procedure, widely adopted in practice, does not explicitly acmunt for the interaction anlong the various stress resultants (shear force, bending monlent and axial force). Nso, the calculatiolls aim to satisfy equilibrium reqoirements, and do not account for the requiremnents of deformation compatibility. In thc absence of shear, liowevcr, the coinbined effect of flexure. and concurrent axial force is made in one step considering the deCorn1ation pattern ('plane section remaining plane'), stress and strain compadbility and equilibrium conditions (refer Chapter 13). Indeed, there have been attempts to design for flexure, shcar and axial force taking all their effects together. The Cosrpresssion Field Tl~cory[Ref. 17.11 is an attempt in this direction. However, the mechanics involved are such that an exact solution is complex and intractable. Hence, recourse has been made to several simplifying assunlptions, which are perhaps questionable. The name "coillprcssion field theory" is based on the analogous problem of the post-buckling shear resistance of thitl-webbed metal girders (plate girders). In such girders, following the buckling of the thin web due to diagonal compression caused by shear, the web cannot msist any more compression. Instcad, the shear is resisted by a 'field of diagonal tension' [Fig. 17.1]. This approach is known as tension field theory. Similarly, in the case of concrete beams, after diagonal cracking, shear would not be resisted by diagonal tension, however a field of diagonal colllpression would still resist shear. This concept came to be called compression field theory.

796

REINFORCED CONCRETE

tension field (diagonal)

SELECTED

DESIGN ,1

\

(a) Seclion

SPECIAL TOPICS 797

.

(bl Princioal compressive stress trajectories

(cl . . Lonaitudinal stra7n

(d) Shear

stress

(a) Longitudinal

stress

Flg.17.1 Tension field in thin-webbed metal girder under shear As in the case of the conventional method dealt with in Chapter 6 , in its simplified version, the comnpltssion fickl theory also uses tlle truss analogy. However, while in the conventional method thc inclination of the diagonal cracks is takcn as 4S0,here, the angle of inclination, 8, of the diagonal compressive stresses is considered variable. Also, thc negative influence of diagonal tension cracking on the diagonal cornprcssive strcngth of concrete [see Seclion 17.1.31 is accounted for. Moreover, thc influence of shear on tlle design of longitudinal reinforcement is accounted for more directly. 17.1.2 General Concepts In onler to understand the complexity involved in an exnct analysislor shear strength, consider the compressive stress trajectories in a beam subjected to a bending moment, M, an axial force, N (considered positive if tensile), and a shcar force, V , as shown in Fig. 17.2. At any section 1-1, the magnitude and dircction of the principal comprcssivc stresses and principal compressive strains will vary over the depth of the section. At the bottom face, the inclination 0 will be 90°, and at the top face 0 will have a minimum value. The shear stress will also vary ovcr the depth of section. On a small element sucli as at A at a dcpth y, the stresses, strains, and tlie con-csponding Mohr's circlcs arc as shown in Fig. 17.2(11) and (j). Concrete is assumed to have no tcnsilc strenglh. In addition, the directions of principal stresses and principal strains are assumctl to coincide. For a comct aaalysis, at each point ovcr the depth of the section, thrco parameters are required to be known/computed. These may be considered, for instance, as the principal strains &I and &1 and the angle 8. h~ addition, the stress-strain relationships for concrete and reinforcing steel are necessary. With these known, the principal stress, fi, can be computed (fl = 0) and hence the normal stress, f,, and the tangential stress, v , at all points over the depth of the scction. The stress ill longitudinal steel,&, can be computed from the stcel strain, &"*, assuming that reinforcing steel caries only axial forces. Thus, the distributions of axial and tangential stresses over the cross section can be obtaincd as shown in Fig. 17.2(e) and (d). By intcgrating these stresses (multiplied by the width of the eross-section) over the depth of scction, the stress resultants N, M and V can be obtaincd [Fig. 17.2(f)]. The slrnin in the transverse direction, E,, determines the tensile stress, f,, in the lransversc shear reinforcement, and the tension in this reinforcement balances the transverse comprcssive stress in the concrete, f,, over the area tributary to it [Fig. 17.2(g)l.

WStress

resultants

(g) Stirrup

(I)

(h) Mohr's circle for stress at A

Mohr's circle for strain at A

Fig.17.2 Stress and strain under combined stress resultants M, N, and V The shrain distribut~onmust be compatible with thc geometry of deformation. Thus, with the usual assumption that plane sections of the beam remain plane (in shallow flexural members), the distributions of E,, q and R must be such that the coi~espondinglongitudinal strain, ex varies linearly over the depth of the member, as shown in Fia. 17.2(c). Obviously, knowinglassuming. a priori, such distributions of E,, E, and Oover the depth at all sections is a tall order! Because of the large number of unknowns involved, a direct solution to the problem is not possible, and a trial and error procedure together with simplifying assumptions has to be used. Two parameters that may be assumed initially are the shear shess distribution and the longitudinal strain distribution. This gives v and E , at all points. Taking trial values of a third parameter also, such as &I,and by successive iterations to satisfy equilibrium, compatibility, and stress-strain relations, the appropriate values o f f , and v at element A can be found. Such a procedure to ~

~

SELECTED SPECIAL TOPICS 799

compute the shear strength of a given section subjected to moment M and axial force N is presented in Ref. 17.2. However, such procedures are lengthy and tedious and seldom resorted to in practice. Instead, approximate procedures are specified in Codes, for example the Canadian specifications CSA A23.3-94. The procedure recommended in CSA A23.3-94 (CI. 11.4) is based on the 'modified compression field theory' [Ref. 17.1, 17.31. This is dealt with in Section 17.1.4.

17.1.3 Stress- Strain Relationship f o r Diagonally Cracked C o n c r e t e For the evaluation of shear strength, the s for steel and concrete must be known. The state of stress Thamaximum (principal) compressive stress in concrete, f,, is inclined at an angle 0 to the axis of the member. The maximum compressive strain alongf, is E,, and the maximum tensile strain, E l , is at right angles to the direction of&. Because of the low tensile strength of concrcte (which is neglected here), tensile cracks will develop early along the direction off,, and the concrete in between these cracks acts as the parallel comprcssion diagonals in the tr68Bnalogy. Therefore, the concrete carrvina . the diagonal compressive strcss has cracks parallel to the direction of compressioil as shown in Fig. 17.3(a). Biaxially strained concrete, as in Rg. 17.3(a), with compression in one direction and a concurrent transverse tensile strain is weaker than concrete in uniaxial compression as in a cube or cylinder test [Fig. I7.3(h)l, where the lateral strain is only due to the Poisson effect. Based on tests [Kef. 17.31 the maximun~compressive strength,&,,,,, of concrete in the preseilce of transverse tensile strain, &I, is given by:

-

f2,,,

< f,' where

Flg.17.3 Stress-strain relationship for diagonally cracked concrete

= f,'/(0.8+170~,)

(17.1)

f ~ = , ~ compressive , strength of concrete in presence of transverse tensile strain f: =specified compressive (cylinder) strength of concrete E, = transverse tensile strain

Eqn. 17.1 give~f,,,,~,,the nraxirmrn strength of concrete under transverse tensile strain 4 . TO compute the stress f2 corresponding to a compressive strain a (concurrent with transverse tensile strain q ) , a stress-strain relation for biaxially strained concrete [Fig. 17.3(a)l is necessary. For this, it may be assumed that the general shape of this stress-strain relation remains the same as for uniaxial compression. One such relationship proposed it1 Ref. 17.4 is given in Eqn. 17.2. Equation 17.2 is also shown in Fig. 17.3(c) where it is compared with the parabolic stress-strain diagram for uniaxial compression

17.1.4 Analysis B a s e d o n Modified C o m p r e s s i o n Field Theory (a) A s s u m p t i o n s and E q u a t i o n s

- C a s e of P u r e S h e a r

To bcgin with, the simple case of a symmetrically reinforced beam under pure shear is considered. The effects of bending moment and axial force are considered subsequently. Prior to cracking, pure shear causes principal tensile and compressive stresses of equal magnitude alo~igdiagonal directions, inclined at 45" to the beam axis. After diagonal tcnsion cracks are fonned, the correspo~~ding tensilc stress in concrete is reduced to zero at the cracks, while the concrete in-betwecn cracks can still sustain tensile stresses. The early compression field theory neglected any contribution to the shear strength from such diagonal tensilc stresses in the cracked concrete and assumed the tensile stress in concrete to be uniformly zem throughout and hence was found to be conservative. In contrast, the rxodifrcd compr.essionfield theory accounts for the contribution of the diagonal tensile stresses in the cracked concrete. Such

800 HEINFORCED

SEL.ECTED SPECIAL TOPICS

CONCRETE DESIGN

801

variation .~~ of tenslle stress in concrete, between cracks, fr ~

tensile stresses vary from zero at the cracks to a ~naximu~n value in between cracks md, for deriving equilibril~mequations, an average value, f , , can be used. This average stress, f,, is lcss than the niaximum tensile stress reached prior to diagonal cracking. Furthermore, the following simplifying assumptions am made in deriving the equations that follow: (i)

.5

r

The shear stress, v, is uniformly distributed over thc web, which has a width b,, and depth 4,(taken as the distance betwccn the resultants of the tensile and compressive forces due to flexure), so that.

v

(17.3) bwdv Under uniform shear stress as above, and with symmetry, the longitudinal (ii) strain, E,, and the inclination, 8, of the principal conipressive stress remain constant over the depth d,, The stress - strain relationshipin compression for the diagonally crackcd (iii) concrete is given by Eqns. 17.1 and 17.2. With these assumptions, the internal forces, stress and straip distributions and the stress resultants at a section subjcctcd to shear only (such :IS at n point of contraflexwc) are as shown in Fig. 17.4. The Mohr's circlcs for strcss and strain states at all points on the section arc shown in Fig. 17.4 (viii) and (ix). From the Mohr's circle of stress. y =-

Y

stresses in symmetri cal

(lit)

unilorm shearstress distrtbution

coi:fe

("1

stress resultants,

(iv) diagonal

~ ~ , , ~ , , , d i , , ~ t("I)

compressive and tensile

stmsser due lo s0car

NvcWfana - fAd"

sheal

Shear

normal

strain +

f'

= ----+ f 2 sill 28 2

-

,

I

v

- fi (17.5) b,,,d, sin 8cos 8 The force in the transverse rcinforcctnent balances the vertical coniponents of the concrete stresses j; and b. Considering equilibrium of stirrup forces and vertical components of f, and fi acting over the concrete area tributary to a stirrup, as shown in Fig. 17.4, A,f, =b,,s(f2sinZ'J- f l cos 2 @ ) (17.6) Substituting forfi from Eqn 17.5.

,

(Mi) Mohr's circle

(vil) Mohr's circle of stress

dlrect strains

of strain

.

,'

I.'

(x) forces in

stirrups

Flg.17.4 Modified compression field theory-Analysisfor shear force V and

v =A,f,d,cote+ S

= v,+ v, where

v,= A " f " 4

f,b,,d, cote

(17.8)

Equation 17,s shows that the shear resistance consists of a part, V,, contributed by the shear reinforcement, and a part, V,, contributed by thc tensile stresses in concrete. The part V, depends on the average tensile stress, f , ,in the diagonally cracked concrek. V, is thc same as derived earlier in Section 6.7.4. (Strictly, the part V, includes a concrcte contribution arising out of the diagonal compressive stressfi also.

SELECTED SPECIAL TOPICS 803

802 REINFORCED CONCRETE DESIGN

The ultimate shear strength in the conventio~lalmethod [Eqn. 6.141 also has a concrete contribution, however its value is derived empirically based on a safe li~nitingvalue for the nominal shear stress in concrete). The stressesfi and fi over a cross section [Fig. 17.4(iv) and (v)] resulting from shear V has a net axial resultant, N,, givcn by: N, = b,, rl, (f2 cosZ@- fi sin%)

(17.11)

There has to be longitudinal reinforcement to resist this. With longitudinal reinforcement symmetrically placed at top and bottom, the tensile force in each of them will be 0.5 N,.. Thus, pure shear necessitates longitudinal reinforcements as well. If A, is the total area of such reinforcement and f, the tensile stress due to shear, A, f, = N,, then substituting forfi from Eqn. 17.5 into Eqn. 17.1 1, A , ~ L =N, = v c o t e

-fi

b,,d,

1'

I

2'

I

(a) Beam loaded in shear

2

1

(17.12)

In Eqns 17.8 and 17.12, f i is the average principal tensilc stress carried by the concrete between diagonal cracks. Based on tests [Ref. 17.31, a relation between average tensile stress, f,, and conesponding average tensilc strain, &,, recommended inRef. 17.1 is: h =&€I for &I 5 & , (17.13)

(c) Calculated average

principal tensile stress,f, where f,, is the tensile stress at cracking and factors a,and account for the bond characteristics of the reinforcement and the type of loading. There are several other considerations in choosing the appropriate value for f,. In deriving the equations above, uniform average stresses and sanirrs have been used. However, the tensilc stress in concrete will be zero at the crack. There will be a corresponding local increase in the tensile stress in the transverse reinforcement, thereby providing the required tensile stress component across the crack interface. Once the stress in the transverse reinforcement (which is highest at the crack location) reaches yield, any increase in shear force can be rcsisted only b y shear stresses, v,, transmitted along the crack interface [Fig. 17.5(b) and (d)]. The magnitude of the shear stress, v,( that can be transmitted between the two sides along the crack interface will depend primarily on the crack width, w , [Fig. 17.5(b)]. The crack width, w, in turn depends on the average tensile strain, &I, and the average spacing, s , of the diagonal cracks. Recommended limiting value of vCito avoid slipping along cracks is [Ref. 17.11:

where, a = the maximum size of aggregate and w may bc taken as thc product of the average principal tensile strain and the average crack spacing so that: w=eIsg

(b) Detail at crack

with diagonal cracks

(17.16)

'(d) Stresses at a crack

Fig.17.5 Transmission of forces across diagonal cracks The spacing of diagonal cracks, sa depends on the type, amount and distribution of the longitudinal and transverse reinforcements. Expressions for estimating ss are given in Ref. 17.1. In Fig. 17.5(c), the average tensile stress in concrete, f,, is assumed to be developed midway between diagonal cracks. As one moves towards the crack, the concretc tensile stress decreases and the slack is taken up by increases in transverse reinforcement stress and/or the interface shear, v, (2V- V,) cot8 Considering the reinforcement on one side only, A,& (V- OSV, ) cot8 are also If stresses due to applied bending moment, M, and axial tension, 4, included, to avoid yielding of the longitudinal reinforcement on the flexural tension side.

i

For members without transverse reinforcements, the spacing of the diagonal racks will be greater than the 300 mm assumed in the above case. For such c a s s also, tables have been prepaed listing 8 and values for vaious cornbinat~onsof longitudinal strain &,and a crack spacing parameter. In both cases, an over estimation of E, will give more conservative predictions of the shear strength.

P

(c) Axial load Fig.17.6 Longitudinal strain at flexural tension steel level

S~te~erary This method aims to ar~iveat morc rational solutions by considering such aspccts as influence of cracking on compressive strcngth, the tensile strengtll of crnckcd concretc, variable angle of inclination of principal stresses, influence of shear on strcsses in longitudi~lalreinforcements, strain compatibility, etc. A1 lllc same time, to make the procedure tractable and suitable for a code format, a series of simplifying assumptions are made. These include neglect of the redistribution of shear stress, assumption of uniform shear stress distribution over the depth, consideration of the stresses and strains at only one level in the cross section and applying the results to the entire section, taking the longitudinal strain at the flexural tension steel level, assunrption of a constant crack spacing of 300 m m for all beams with shear reinforcement. assumption that ihc shear reinforcement yields, working out the equations for pure shear and accountmg for effects of flexure and axial load by modifying longitudinal steel strain only

.

..

Despite the above simplifying assun~ptions,a closed form solution is not possible and a trial-and-error approach involving complex equations, tables and charts are used. Even so, modification factors have to be applied in some cases, as with the

808 REINFORCED CONCRETE DESIGN

constant 1.3 in Eqn. 17.33. Unfortunately, the objective for rigour is compromised by the need to introduce so many assumptions. Indeed, the traditional method uses far less and more justifiable assumptions and is supported by the sound concepts of the Strut-and-Tie model. In any case, IS Code does not specify Compression Field Theory as a method for design in shear. For these rcasons, the authors do not recommend this a s a standard method for design especially in the Indian context. However, the topic has been included as it is a relatively recent theoretical development. For the sake of cotnpleteness, the CSA Code provisions are given below and an example workcd out later [Ex. 17.11.

17.1.6 CSA Code Provislons for Shear Design by the Compression Field Theory IS 456:2000 does not include any provision based on compression field theory. One of the Codeswhich introduced the comoression field theorv for shear desien earlv on is the Canadian Standards Association (CSA) Standard CSA A23.3-94: Design of Concrete Str.nclrr,rs. The provisions in that Code are briefly discussed here. CSA Standard uses the cylinder strength- fc' as.the specified concrete strength. To relate this to the cube strength, Eqn. 2.3 may be used. Morcaver, this Code uses material resistaxe furrors of & = 0.6 for concrete and = 0.85 for reinforcing bars. These material resistance factors correspond to the inverse of the i~anialsafety factors for materials, y, referred to in Section 3.6.2. The factor A accounts for the effects of concrete density on tensile strength and other properties: [see also Section 6.9.21. The procedure designated as "general ntethos' for shear design in CSA A23.3-94 (CI. 11.4) follows the simplified procedure described in Section 17.1.5. The load and resistance factors are also incorporated. The controlli~~g dcsign etquation is:

-

Vrg = VCg+ VW 2 VI where,

Vrg V, V, V,

A

ensure that the transverse reinforcement will yield prior to the crushing of the concrete in the weo in diagonal compression, V , is limited to: (17.36) v,, ~ 0 . W , f c ' b , d , Tables and graphs are presented in the Code for determining values o f 0 and Ofor sections with and without the minimum amount of transverse reinforcements For sections with transverse reinforcement, the table is in terms o f parameters v,/(A r$* f,'), where ",is the factored shear stress, and E,, the longitudinal strain at the tension steel level. For evaluating these parameters, (17.37) v = V, l(b,,d,) and E,

=[0.5(Nf t V , c o t 8 ) + M f /d,jl(E,A,)

(17.38)

< 0.002 For sections without the minimum transverse reinforcement, the parameters to be used are the crack spacing parameter, s,, determination of which is as per CSA Code C1.11.4.7, and E,. Longitudinal reiuforcement is to be designed for the combined effects of flexure. axial load and shear. Accordingly, as in Eqn. 17.31, at all sections, (17.39) AJ, Z M l d~, + 0 . 5 ~ /+(v1 - 0 . 5 ~ ~ ) c o t e

(17.32)

is the factored shear resistance, is the factored shear resistance attributed to concrcte, is the factored shear resistance provided by the sbcar reinforcement, is thc factored shear force at the section, and is the factor to account for low density concrcte

vw = 1 . 3 ~ ~ c ~ v E b , 9 d v

(17.33)

For stkrups perpendicular to beam axis:

v,

=

A,4,fYd, cots S

(17.34)

For transverse reinforcement inclined at an anele a lo the loneitudinal axis. A,$, f#,(cot .9+ cot a)sina VQ = (17.35)

Factored shear force

I

Design shear (average)

Fig. 17.7 Design for average shear over lenglh &cot0

S

The factor 1.3 in Eqn. 17.33 compensates for the low valuc of & and partially offsets the co~~servntistn of this method. The expnxsions for V,, are the same as those derived on the basis of the truss model in Section 6.7.4 [Eqns. 6.18(a)]. To

In the case of members not subjected to significant axial tension, the ~equirelllent of Eqn. 17.39 may be satisfied by extending the flexural tension reinfo~cenlenta distance of d, cot0 beyond the location needed for flexure alone [compare with Fig. 6,10(a)j. Similarly computation similar to development of Eqn. 6.21 [Fig. 6.10(b)J

SELECTED SPECIAL TOPICS 81 1

810 REINFORCED CONCRETE DESIGN

with di;lg~~llal crdck at ;illgle O will show lhnt at exterior direct bearing seppurts, the homm lo~~giludin.~l remforccl~~m sl~ouldbz u m ~ h l rc~frcsirtine a tcnulc iolcc T ; 146 kN, spacing limits are 300 mm and 0.35d Design of stirrups

The factored shear resistance contributed by concrete is [Eqn. 17.331 V, = l . 3 x l.OxO.6xO.1OOX f i x 3 0 0 x 3 9 1 X I O " = 40.9 kN The factored shear resistance to be psovided by stirlnps is

1

Assumiug No. 10 U stinups placed perpcntlicula~.lo beam axis, the required spacing is givcn by [Eqn. 17.341 s =($,A,.f,d,,cotO)N,, = (0.85 x 200 x 400 x 391 cot 35") 1 189 100 = 201 mm As ,fI > 146 kN,llle limiling spacing is given by 300 111111or 0.35 d = 0.35 x434 = 152 mln

I

Hcncc, the limiting spacing controls, and a spacing of 150 m n is selected. The shear force is 146 kN at a distance of 1970 mm from the face of support, and the limiting spacing is applicable upto this location. Therefore, provide the first stirrup at a distance of 75 m n from the face of the support, followed by 13 more stirrups at 150 nun, covering a total length of 2025 mm from the face of the support.

1

.

Critical section at d, = 391 mm from face (or 371+120 = 511mn from centre) of support, where Vj= 230 W. The parameters 0 anti 0 are to be determined from Table 11.1 of CSA A23.3.94. For this the factored shear stress is v/ = Vf/(b,, d,) = 230 x 10 l(300 x391) = 1.96 MPa Factored shear stress ratio, v, /(A r&&') = 1.96 l(1.0 x 0.6 x 20) = 0.163 Longitudinal strainis given by Eqn. 17.38, E, = ( 0.5 Vfcol0 + Mf/d, ) 1 (E,Ac), where Mj is the bending moment at the critical section at 391 mm from the support, cotresponding to the load causing maximum shear at this section. However, for convenience and to be conservative, this moment is taken here as the moment with full load on the entire span. Accordingly, M, = 6 4 . 2 x ( 4 x 0 . 5 1 1 - 0 . 5 1 1 ~ 1 2 ) = 122.8 kN.m E, = (0.5 x 230 x 1 0 ~ c o t 0+ 122.8 X l o 6 / 391) l(200 000 x 2100) = ( 0.274 cot0 + 0.748 ) x

i

!

I

1

Section at 2.3 m From face of support The Code pcrmits the design of stirrups for the average shear ovcr a length of d, cot8. Here, rl, cot0 is in the range of 39kot35" = 558 nun (although both d, and 8 will vary slightly along the span) In practice, designing for every, discrete lengths of d, cot0 is not warranted. In this example, the next section for design is taken at a distance of 2 m from face of support (which is approximately 55812 nnn from the location of the last stirrup designed. The shear at 2.3 m from face (or 2.42 m from centre) of support is V, = 44.6 + (257 - 44.6) (4 -2.42) I 4 = 128.5 kN. The bending momcut corresponding to this shear is, M,=19.6x4x2.42-19.6x2.422/2+44.6x5.582x2.42/(8x2) = 342.4 kNm.

820

St1 ECTED SPECIAL TOPICS

REINFORCED CONCRETE DESIGN

At 2.3 m from face or support, there arc 5 No. 30 bars and effective depth is diffcrent from at support. Here, conservatively, the cffcctivc depth at mid-span, cqual to 406 m m will bc used for this section also. Conesponding (1, = 0.9 x 406 = 365 mm. Factored slrcar stress mtio, v,Xh & f,') = 128.5 x lo3/(300 x 365 x 1.0 x 0.6 x 20) = 0.098 Longitudinal strain is E,, = (0.5 x 128.5 x lo3 cot0+ 342.4 x 106/365)/ (200 000 x 3500) .- (0.092 cot0 4- 1.340) x 10' For !?/(A q5 x ,,, (not permitted in design).

835

836

REINFORCED CONCRETE DESIGN

Table A.2(b) Contd

Table A.Z(b) Contd

APPENDIX A

839

838 REINFORCED CONCRETE DESIGN Table A.2(b) Contd

Table A.3(a) DESIGN AIDS (LSM) for singly reinforced rectangular beam sections Values of p, for given values of R E M J ~ ~ ( M P ~ )

(M 20, M 25 concrete grades)

Note: Values below the horizontal line in each column correspond to the condition xu > x ,,,,, (not permitted in design). -[refer Section 4.1.21.

840

APPENDIX A

REINFORCED CONCRETE DESIGN

Table A.3(a) Contd

Table &?.(a) Contd

* p , > P , , , ~,~

not admissible in design [refer Section 57.21

841

APPENDIX A

842 REINFORCED CONCRETE DESIGN Table A 4 a ) Contd

Table A.3(b) DESIGN AIDS (LSM) for singly reinforced rectangular beam sections Values of pt for given values of R 2 ~ulb#(MPa) (M 30, M 35 concn

* p, > P , ~ , , ,, not admissible in design [refer Section 57.21

843

grades)

848

REINFORCED CONCRETE DESIGN

Table A.3(b) Contd

* p, > P , , , ~, ~ not admissible in design [refer Section 4.7.21

APPENDIX A

849

Table A.4a DESIGN AIDS (LSM) for doubly reinforced rectangular beam sections Values of id p,for given values of R - M J ~ ~ ( M P ~ ) 'e 415 steel. M 20 concrete)

APPENDIX A

854

855

REINFORCED CONCRETE DESIGN

Table A.4a (Fe 415 steel, M 20 concrete) Contd

Table A.4b DESIGN AIDS (LSM) for doubly reinforced rectangular beam sections Values of 7d p,for given values of R - M J ~ ~ ? ( M P ~ ) Fe 415 steel, M 25 concrete)

856

REINFORCED CONCRETE OESIGN

Table A.4b (Fe 415 steel, M 25 concrete) Contd

APPENDIX A

Table A.4b (Fe 415 steel, M 25 concrete) Contd

857

APPENDIX A Table A.4b (Fe 415 steel, M 25 concrete) Contd

Table A.5 Areas (rnm2)of reinforcing bar groups

861

862

REINFORCED CONCRETE DESIGN

Table A.6 Areas of uniformly spaced bars in slabs, in mm2/m

r

i

Table 8.1 DEAD LOADS - Unit Weights of Some Materials/Components Table 8.2 LlVE LOADS on Floors Table 8.3 LlVE LOADS on Roofs Table 8.4 HORIZONTAL LlVE LOADS on Parapets/Balustrades

D

E f

C

APPENDIX B 865

864 REINFORCED CONCRETE DESIGN Table 8.1 DEAD LOADS

Table 8.2 LlVE LOADS o n Floors

- U n ~Weights t of Some MaterialslComponents

.

Residential Office - with separate storage

Concrete

- plain

Concrete - reinforced

24

-without separate storage

25

Shops, Classrooms, Waiting rooms, Restaurants. Work rooms, Theatres, etc.

Brick masonry

19-20

Stone masonry

21-27

Timber

6-10

Piaster- cement Plaster

- lime

Steel

21 18 78.5

-with fixed seating

.

- without fixed seating Factories and Warehouses Stack rooms In Libraries, Bookstores Garages

- h e a v vehicles

Roofing - AC sheet Roofing - GI sheet

- light vehicles

stairs', Landings, Balconies

Mangalore tiles with batiens

-not

liable to overcrowding

Roof finishes

-liable

to overcrowdlno

Floor finishes Steel work lor roofing

I;

Table 8.3 LIVE LOADS o n Roofs

Roof with access Roof without access Sloping roofs (e > lo0)

oi.55

0.75 - 0.02 x (8

- loo)

Table 8 4 HORIZONTAL LlVE LOADS o n ParapetsIBalustrades

Places of assembly

2.25

Others

0.75

I n thc case of cantilever steps, a minimum of 1.3 !di concentrated load (at the free edge) should be supported by each cantilever step. assumed to act horizontally on rap of parapethalustrade.

additives, 36 standards, 71 Aggregate, grading requirements, 31 properties and tests, 30 standards, 71 types of, 30 Aggregate interlock, 230 Allowable @erl~ssible)stresses, 115 Analysis, 21,95, 169, 189 Anchorage, 299,306,760 Arch action, 303 Areas of reinforcing bars, 184, 861 Axial compression, 586 strength, 42, 590 tension, 19

section, 109,116,136,159 strain condition, 136. 597 Bars, areas of, 184, 861, 862 bending of, 219 bundled, 172,219,301 curtailrncnt, 210-2 14 cut off, 210 hanger, 128,237 hooks, 306 spacing, 172, 862 sizes, 66

Beam, analysis, 95 aids, 121, 144, 8 2 9 4 3 8 compression reinforcement. 128, 153 L- and T-sections, 122.. 129.. 147 rectangular section, 112, 128, 181 service loads, 112 slab modelled as beam, 160 ultimate loads, 134 balanced failure, 109, 136 balanced section, 109, 116, 136, 159 balanced stain condition, 136 bm spacing, 172 behaviour in flexure, 95-168 bond stesses, (see Bond) continuous, 189, 324 cover, 170 crackcontrol. 172.. 357.. 391 deep, 180 deflection control, 176,357-388 design, 169-224 aids, 184, 199, 839-862 conlpression reinforcement, 197 L- and T- sections, 203 rectangular section, 181, 197 tension reinforcement, 183 doubly reinforced, 128, 153, 197 effective depth. 110 elastic theory, 99

868 INDEX

'/

flexural behavioor, 105-112 failure, 110 moment coefficients, 189,324 moment-curvaIurc. 110 over-reinforced, 109 over-rcinrorced (WSM), 119 primary, 13,269 secondary. 13,269 servicc load stresses, 112 shear strength, (see Shcar) singly reinforced, 112, 137, 181 size guidelines, 179, 181 steel percentages. maximum, 175 minimum, 174 stress block, 98, 137 T-beam analysis, 122, 147 T-beam d e s i g n 203 torsion, 267-294 transformed section, 101 under-reinforced. 116 under-reinforced (WSM), 109 Bearing capacity, 659 pressure, 660 stress, 671 Biaxial bending with compression, 625634 eccentricity, 567 Bond, anchorage, 299 bundled bars, 172,219,301 code requirements, 305 compression bars, 299, 301 development, 299, 306 development length, 213. 300 failure, 301 flexural, 297,305 hook, 306 mechanisms, 295 splices, 308 stress, 296 tests, 303 Braced frame, 572

INDEX

Building failules, 751 Buildings, ~eioforcedconcrete, 9-20 Bundled bars, 172,219,301 Buttress wall, 705

Cantilever wall, 705 Capital. 466 Carry-over factor, 484 Cement, hydration or, 26 paste, 26 tests on, 29 standards, 71 types of, 26-29 Chair, 764 Characteristic load and strength, 87 Codes and handbooks, 22 Coefficient of thcrmal expansion, 59 of static friction, 710 Column, additional moment, 640 axial con~pression.586-593 balanccd strain condition, 597 braced, 572, 636 capital, 466 circular, 566, 620 compression failore, 597 dcsign aids, 618 eccentric loading, uniaxial, 594-624 biaxial, 625-634 effective length, 569-573 eccentricity, minimum, 582 equivalent, 488 interaction curve, diagram. 597, 610,618 interaction surface, 627 latcml drift elfect, 637 load contour, 628 mcmber stability effect. 637 moment magnification, 640 P-A cffect, 634,636,637 plastic centroid, 587

reinforcerncnt requirements, 582 short, 568, 586 service load behaviour, 587 ultimate load behaviour,. 588 slender (or long), 568, 6 3 6 6 4 8 behaviour, 634 design, 639 slenderness l i t ~ t s 581 , ratios, 568 stability effect, 636 spiral, 565 tension failure, 597 tied, 565 types, 565-568 unsupported length, 569 Column sJ~ip,460 Combined bending and axial load, 594,625 Combined shear and torsion, 282 Combincd footing, 692-702 Compaction, 33 Compatibility torsion, 267 Composite material, 5, 566 Compression failure, 111 Compression field theory 795 Compression member, (see C o l u m ) Compression reinforcement, 122, 153,582 Concrete, aerated, 38 admixtures, 37 age effect, 50 bleeding, 33 brittle nature, 5 characteristic strength. 39 chemical attack, 63 compressive strength, axial, 42 biaxial, 53 cube, 42 cylinder, 42 flexural, triaxial, 55 coefficient of thermal expansion,

869

59 combined stresses, 53 compaction, 3 3 confinement, 55, 586,785 constituents, 4, 25-61 cover, 61, 171 creep, 55 curing, 35 ductility, 5, 55 durability, 5 9 fibre-reinforced, 7 ' grade, 38 mix design, 4 0 modulus of elasticity, 46 nominal mix, 40 permeabilily, 6 3 porosity, 63 plain, 4 Poisson's ratio, 47 prestressed, 7 reinforced, 4 segregation, 31, 33 shear strength, 53 sMnkage, 57 softening, 46 standards, 70 stress block, 137 stress-strain curves, 44, 52, 89 technology, 25 temperature effects, 57 tensile strength, cylinder splitting, 5 2 modulus of rupture, 51 ultimate (crushing) strain, 44, 89 water content, 33 water cement ratio, 33 workability, 33 Connection details, 238,754 Construction, continuity in, 317 joints, 764 phascs in, 8 practices, 766 Corner reinforcement, 420,432 Corrosion, 65

870

INDEX 871

INDEX

Counterfort wall, 705 Cover, 65, 170,764 Cracking, control of, 65,391 moment, 51, 101 Cracked second moment of area, 113 section, 107 Cracked-transformed section, 113 Crackwidth, 391 Creep, 55,384 Curvature, 110, 334, 381,385,772

Dcflection, allowable, 176,358 control of, 176,388,765 creep, 384 elastic theory, 360 immediate, 360 long-term, 176 modification factors, 177 shrinkage, 381 temperature effects, 387 Design aids. 184. 199,618, 839-862 for bond, '295-316 of beams, 169-2'24 of columns, 565-654 drawings, 754 earthquake-resistant, 771-792 mix concrete, 41 objectives, 7 of footings, 655-702 for flexure, 169-224 philosophies, LSM, 78 probabilistic, 74 ;eliability, 7,6 ULM. 80 WSM, 79 of retaining walls, 703-744 for shear, 225-226 of slabs,

flexural rigidity, 361 span, 192 Elastic analysis, 99, 317, 324, 422 behaviour, 107, 228, 271, 587 modulus, 46, 68 End-bearing splice, 310 Equilibrium torsion, 267 Equivalent frame method, 481-493

one-way, 169-22.4 two-way, 417-532 of staircases, 533 -564 fottorsion, 267-294 Detailing, 749-767 Development length ,213,300 Diagonal tension, 214,229 Differential settlen~ent,655 Differential shrinkage, 58,381 Direct design method, 4 6 9 4 8 1 Direct tension, 5 Distributors, 175 Doubly reinforced beams, 122, 153, 197 Dowel action. 229 bars, 673 D r o vanel. ~ 465 Ductile failure, 109,231,772 frame, 776,780-788 wall, 776, 789 Ductility, 5, 55,772 Durability, 59 Dynamic modulus. 46

-.

Earthauake load. 9 ~> Earthquake-resistant design, 771-792 columns, 785 design considerations, 778 ductility, 772-777 flexural members, 791 foundations, 790 joints in frames, 788 materials, 790 walls, 789 Economy, 7 Edge beam, 476 Edge strip, 428 Effective curvature, 364 depth, 110 flange width, 123 length, 568

'

Factored load, 140 moment, 140, 181 shear force, 234 soil pressure, 665 torque, 275 Failure surface, 627 Ferrocement, 7 Rbre-reinforced concrete. 7 R r e resistance 822 Fixed-end nlomenl, 484 Flange, 122 Flanged sections, 122, 129, 147 Flat plate, 15 Flat slab, 16 Flexural behavioor, 95-168 Flexural tcnsion, 5 Floor systems. I I Footing, 655-702 allowable soil pressure, 659 bearing at column base, 671 code requirements, 665 combined, 658,692-702 design considerations, 665 dowels, 673 isolated, 658 pedestal, 673 net soil pressure, 659 overturning, 664 shcar key, 665 '~ sliding, 664 '\ trapezoidal, 693 lypes of, 656

pull-out, 665 wall, 666 Fran~canalysis, app~oxilnations,317, 324,456 continuity, 317-341 procedures, 324 sliffness of members, 3 2 4 Friction, 710

of concrete, 38 of steel, 66. I36 Gravity loads, 9, 321, 328 Grid floor. 13

Hooks, 306 Hooo reinforcement. 784 Hydration, 26 Hysterjsis, 46

...

*- *,.~ ,;;>;-:;~.:~.